Textbook - Triangles Questions in English

(4:1) Since $AB \parallel CD,$
$\therefore \angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$ (Alternate interior angles).
In $\triangle AOB$ and $\triangle COD,$
$\angle AOB = \angle COD$ (Vertically opposite angles).
$\angle OAB = \angle OCD$ (Alternate interior angles).
$\angle OBA = \angle ODC$ (Alternate interior angles).
$\therefore \triangle AOB \sim \triangle COD$ (By $AAA$ similarity criterion).
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\triangle AOB)}{\operatorname{ar}(\triangle COD)} = \left(\frac{AB}{CD}\right)^2$.
Given $AB = 2 CD,$
$\therefore \frac{\operatorname{ar}(\triangle AOB)}{\operatorname{ar}(\triangle COD)} = \left(\frac{2 CD}{CD}\right)^2 = \left(\frac{2}{1}\right)^2 = \frac{4}{1} = 4:1$.
Thus,the ratio of the areas of triangles $AOB$ and $COD$ is $4:1$.

(N/A) Let us draw two perpendiculars $AP$ and $DM$ on line $BC$.
We know that the area of a triangle $= \frac{1}{2} \times \text{Base} \times \text{Height}$.
Therefore,$\frac{\operatorname{ar}(\Delta ABC)}{\operatorname{ar}(\Delta DBC)} = \frac{\frac{1}{2} \times BC \times AP}{\frac{1}{2} \times BC \times DM} = \frac{AP}{DM}$.
In $\triangle APO$ and $\triangle DMO$:
$\angle APO = \angle DMO = 90^{\circ}$ (By construction).
$\angle AOP = \angle DOM$ (Vertically opposite angles).
Therefore,$\triangle APO \sim \triangle DMO$ (By $AA$ similarity criterion).
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{AP}{DM} = \frac{AO}{DO}$.
Substituting this into our area ratio equation:
$\frac{\operatorname{ar}(\Delta ABC)}{\operatorname{ar}(\Delta DBC)} = \frac{AO}{DO}$.

(N/A) Let us assume two similar triangles as $\triangle ABC \sim \triangle PQR$.
According to the theorem of areas of similar triangles:
$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{AC}{PR}\right)^2$ $...(1)$
Given that,$\text{ar}(\triangle ABC) = \text{ar}(\triangle PQR)$.
Therefore,$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)} = 1$.
Substituting this value in equation $(1)$,we get:
$1 = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{AC}{PR}\right)^2$.
This implies:
$\left(\frac{AB}{PQ}\right)^2 = 1 \Rightarrow AB^2 = PQ^2 \Rightarrow AB = PQ$
$\left(\frac{BC}{QR}\right)^2 = 1 \Rightarrow BC^2 = QR^2 \Rightarrow BC = QR$
$\left(\frac{AC}{PR}\right)^2 = 1 \Rightarrow AC^2 = PR^2 \Rightarrow AC = PR$
Since all three corresponding sides are equal,by the $SSS$ (Side-Side-Side) congruence criterion,$\triangle ABC \cong \triangle PQR$.

(1:4) Given that $D, E$ and $F$ are the mid-points of sides $AB, BC$ and $CA$ respectively in $\Delta ABC$.
By the Mid-point Theorem,the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.
Therefore,$DE \parallel AC$ and $DE = \frac{1}{2} AC$.
Similarly,$EF \parallel AB$ and $EF = \frac{1}{2} AB$,and $DF \parallel BC$ and $DF = \frac{1}{2} BC$.
In $\Delta DEF$ and $\Delta ABC$:
$\frac{DE}{AC} = \frac{EF}{AB} = \frac{DF}{BC} = \frac{1}{2}$.
By $SSS$ similarity criterion,$\Delta DEF \sim \Delta ABC$.
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\operatorname{ar}(\Delta DEF)}{\operatorname{ar}(\Delta ABC)} = \left(\frac{DE}{AC}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Thus,the ratio of the areas of $\Delta DEF$ and $\Delta ABC$ is $1:4$.

(N/A) Let us assume two similar triangles $\Delta ABC \sim \Delta PQR$. Let $AD$ and $PS$ be the medians of these triangles.
Since $\Delta ABC \sim \Delta PQR$,we have:
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \quad ...(1)$
And $\angle B = \angle Q \quad ...(2)$
Since $AD$ and $PS$ are medians,$D$ and $S$ are midpoints of $BC$ and $QR$ respectively.
Therefore,$BD = \frac{BC}{2}$ and $QS = \frac{QR}{2}$.
Substituting these into equation $(1)$:
$\frac{AB}{PQ} = \frac{2BD}{2QS} = \frac{BD}{QS} \quad ...(3)$
In $\triangle ABD$ and $\triangle PQS$:
$\angle B = \angle Q$ [From $(2)$]
$\frac{AB}{PQ} = \frac{BD}{QS}$ [From $(3)$]
Therefore,$\triangle ABD \sim \triangle PQS$ by $SAS$ similarity criterion.
This implies that $\frac{AB}{PQ} = \frac{AD}{PS} \quad ...(4)$
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \left(\frac{AB}{PQ}\right)^2$
From equation $(4)$,substituting $\frac{AB}{PQ} = \frac{AD}{PS}$:
$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \left(\frac{AD}{PS}\right)^2$
Hence proved.

(N/A) Let $ABCD$ be a square of side $a$.
Therefore,its diagonal $= \sqrt{2} a$.
Two equilateral triangles are formed: $\Delta ABE$ (on side $AB$) and $\Delta DBF$ (on diagonal $DB$).
Side of equilateral triangle $\Delta ABE$ described on side $AB = a$.
Side of equilateral triangle $\Delta DBF$ described on diagonal $DB = \sqrt{2} a$.
Since all equilateral triangles are similar,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
$\frac{\text{Area of } \Delta ABE}{\text{Area of } \Delta DBF} = \left( \frac{a}{\sqrt{2} a} \right)^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}$.
Thus,$\text{Area of } \Delta ABE = \frac{1}{2} \times \text{Area of } \Delta DBF$.

(N/A) In $\Delta ACD$ and $\Delta ABC$:
$\angle ADC = \angle ACB = 90^{\circ}$
$\angle CAD = \angle CAB$ (Common angle)
Therefore,$\Delta ACD \sim \Delta ABC$ (by $AA$ similarity criterion).
This implies $\frac{AC}{AB} = \frac{AD}{AC}$,so $AC^{2} = AB \cdot AD$ $...(1)$
In $\Delta BCD$ and $\Delta BAC$:
$\angle BDC = \angle BCA = 90^{\circ}$
$\angle CBD = \angle ABC$ (Common angle)
Therefore,$\Delta BCD \sim \Delta BAC$ (by $AA$ similarity criterion).
This implies $\frac{BC}{BA} = \frac{BD}{BC}$,so $BC^{2} = BA \cdot BD$ $...(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{BC^{2}}{AC^{2}} = \frac{BA \cdot BD}{AB \cdot AD}$
Since $BA = AB$,we get:
$\frac{BC^{2}}{AC^{2}} = \frac{BD}{AD}$

(B) Let $AB$ be the ladder and $CA$ be the wall with the window at $A$. The wall and the ground form a right angle at $C$. Thus,$\triangle ABC$ is a right-angled triangle with $\angle C = 90^\circ$.
Given: $BC = 2.5\, m$ (distance of the foot of the ladder from the wall) and $CA = 6\, m$ (height of the window from the ground).
According to the Pythagoras Theorem,in a right-angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides:
$AB^2 = BC^2 + CA^2$
$AB^2 = (2.5)^2 + (6)^2$
$AB^2 = 6.25 + 36$
$AB^2 = 42.25$
$AB = \sqrt{42.25}$
$AB = 6.5\, m$
Therefore,the length of the ladder is $6.5\, m$.

(N/A) Given: In $\Delta ABC$,$AD \perp BC$.
To prove: $AB^2 + CD^2 = BD^2 + AC^2$.
Proof:
In right-angled $\Delta ADB$,by Pythagoras theorem:
$AB^2 = AD^2 + BD^2$ --- $(1)$
In right-angled $\Delta ADC$,by Pythagoras theorem:
$AC^2 = AD^2 + CD^2$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$AB^2 - AC^2 = (AD^2 + BD^2) - (AD^2 + CD^2)$
$AB^2 - AC^2 = AD^2 + BD^2 - AD^2 - CD^2$
$AB^2 - AC^2 = BD^2 - CD^2$
Rearranging the terms,we get:
$AB^2 + CD^2 = BD^2 + AC^2$
Hence proved.

(N/A) $BL$ and $CM$ are medians of $\Delta ABC$ in which $\angle A = 90^{\circ}$.
From $\Delta ABC$,by Pythagoras theorem:
$BC^2 = AB^2 + AC^2$ $...(1)$
From $\Delta ABL$,since $L$ is the midpoint of $AC$,$AL = AC/2$:
$BL^2 = AB^2 + AL^2 = AB^2 + (AC/2)^2 = AB^2 + AC^2/4$
$4BL^2 = 4AB^2 + AC^2$ $...(2)$
From $\Delta CMA$,since $M$ is the midpoint of $AB$,$AM = AB/2$:
$CM^2 = AC^2 + AM^2 = AC^2 + (AB/2)^2 = AC^2 + AB^2/4$
$4CM^2 = 4AC^2 + AB^2$ $...(3)$
Adding $(2)$ and $(3)$:
$4(BL^2 + CM^2) = 4AB^2 + AC^2 + 4AC^2 + AB^2$
$4(BL^2 + CM^2) = 5AB^2 + 5AC^2$
$4(BL^2 + CM^2) = 5(AB^2 + AC^2)$
Using equation $(1)$,$AB^2 + AC^2 = BC^2$,so:
$4(BL^2 + CM^2) = 5BC^2$.

(N/A) Through $O$,draw $PQ \parallel BC$ such that $P$ lies on $AB$ and $Q$ lies on $DC$.
Now,$PQ \parallel BC$.
Therefore,$PQ \perp AB$ and $PQ \perp DC$ (since $\angle B = 90^{\circ}$ and $\angle C = 90^{\circ}$).
So,$\angle BPQ = 90^{\circ}$ and $\angle CQP = 90^{\circ}$.
Therefore,$BPQC$ and $APQD$ are both rectangles.
Now,from $\Delta OPB$,by Pythagoras theorem:
$OB^{2} = BP^{2} + OP^{2}$ $...(1)$
Similarly,from $\Delta OQD$:
$OD^{2} = OQ^{2} + DQ^{2}$ $...(2)$
From $\Delta OQC$,we have:
$OC^{2} = OQ^{2} + CQ^{2}$ $...(3)$
And from $\Delta OAP$,we have:
$OA^{2} = AP^{2} + OP^{2}$ $...(4)$
Adding $(1)$ and $(2)$:
$OB^{2} + OD^{2} = BP^{2} + OP^{2} + OQ^{2} + DQ^{2}$
Since $BP = CQ$ and $DQ = AP$ (opposite sides of rectangles $BPQC$ and $APQD$):
$OB^{2} + OD^{2} = CQ^{2} + OP^{2} + OQ^{2} + AP^{2}$
Rearranging the terms:
$OB^{2} + OD^{2} = (CQ^{2} + OQ^{2}) + (AP^{2} + OP^{2})$
Using equations $(3)$ and $(4)$:
$OB^{2} + OD^{2} = OC^{2} + OA^{2}$.

(B) We know that all equilateral triangles have all their angles equal to $60^{\circ}$ and all their sides of the same length. Therefore,all equilateral triangles are similar to each other.
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Let the side length of $\triangle ABC$ be $x$.
Since $D$ is the mid-point of $BC$,the side length of $\triangle BDE$ is $\frac{x}{2}$.
Therefore,the ratio of the areas is:
$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle BDE)} = \left(\frac{x}{\frac{x}{2}}\right)^{2} = (2)^{2} = \frac{4}{1}$.
Hence,the ratio is $4:1$,and the correct answer is $(B)$.

(C) If two triangles are similar to each other,then the ratio of the areas of these triangles is equal to the square of the ratio of their corresponding sides.
Given that the ratio of the sides is $4:9$.
Therefore,the ratio of the areas of these triangles $= (4/9)^2 = 16/81$.
Hence,the correct answer is $(C)$.

(A) The given sides of the triangle are $7 \, cm, 24 \, cm,$ and $25 \, cm$.
To check if it is a right-angled triangle,we use the converse of the Pythagoras theorem,which states that if the square of the longest side is equal to the sum of the squares of the other two sides,then the triangle is a right-angled triangle.
First,we calculate the squares of the given sides:
$7^2 = 49$
$24^2 = 576$
$25^2 = 625$
Now,we check if the sum of the squares of the two smaller sides equals the square of the largest side:
$49 + 576 = 625$
Since $7^2 + 24^2 = 25^2$,the given sides satisfy the Pythagoras theorem.
Therefore,the triangle is a right-angled triangle.
The longest side of a right-angled triangle is its hypotenuse. Thus,the length of the hypotenuse is $25 \, cm$.

(N/A) The given sides of the triangle are $3 \text{ cm}, 8 \text{ cm},$ and $6 \text{ cm}$.
According to the converse of the Pythagoras theorem,a triangle is a right triangle if the sum of the squares of the two smaller sides is equal to the square of the longest side.
Here,the squares of the sides are:
$3^2 = 9$
$6^2 = 36$
$8^2 = 64$
Checking the sum of the squares of the two smaller sides:
$3^2 + 6^2 = 9 + 36 = 45$
Since $45 \neq 64$ (i.e.,$3^2 + 6^2 \neq 8^2$),the condition for a right triangle is not satisfied.
Therefore,the given triangle is not a right triangle.

(NONE) Given that the sides of the triangle are $50 \text{ cm}, 80 \text{ cm},$ and $100 \text{ cm}.$
To check if it is a right-angled triangle,we apply the converse of the Pythagoras theorem,which states that in a triangle,if the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides,then it is a right-angled triangle.
Calculate the squares of the sides:
$50^2 = 2500$
$80^2 = 6400$
$100^2 = 10000$
Now,check the sum of the squares of the two smaller sides:
$50^2 + 80^2 = 2500 + 6400 = 8900$
Comparing this with the square of the longest side:
$8900 \neq 10000$
Since $50^2 + 80^2 \neq 100^2$,the given triangle does not satisfy the Pythagoras theorem.
Therefore,the given triangle is not a right-angled triangle.

(A) Given that the sides are $13\, cm, 12\, cm,$ and $5\, cm$.
Squaring the lengths of these sides,we obtain $169, 144,$ and $25$.
Clearly,$144 + 25 = 169$.
Or,$12^2 + 5^2 = 13^2$.
The sides of the given triangle satisfy the Pythagoras theorem.
Therefore,it is a right triangle.
We know that the longest side of a right triangle is the hypotenuse.
Therefore,the length of the hypotenuse of this triangle is $13\, cm$.

(N/A) Let $\angle MPR = x$.
In $\Delta MPR$,
$\angle MRP = 180^{\circ} - 90^{\circ} - x = 90^{\circ} - x$.
Similarly,in $\Delta MPQ$,
$\angle MPQ = 90^{\circ} - \angle MPR = 90^{\circ} - x$.
In $\Delta MPQ$,
$\angle MQP = 180^{\circ} - 90^{\circ} - (90^{\circ} - x) = x$.
Now,in $\Delta QMP$ and $\Delta PMR$:
$\angle MPQ = \angle MRP = 90^{\circ} - x$
$\angle MQP = \angle MPR = x$
$\angle PMQ = \angle RMP = 90^{\circ}$
Therefore,$\Delta QMP \sim \Delta PMR$ (by $AA$ similarity criterion).
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{QM}{PM} = \frac{PM}{MR}$
$\Rightarrow PM^{2} = QM \cdot MR$.

(N/A) Given: In $\triangle ABD$,$\angle BAD = 90^\circ$ and $AC \perp BD$.
To prove: $AB^2 = BC \cdot BD$.
Proof: Consider $\triangle BCA$ and $\triangle BAD$.
In $\triangle BCA$ and $\triangle BAD$:
$\angle BCA = \angle BAD = 90^\circ$ (Given $AC \perp BD$ and $\angle BAD = 90^\circ$)
$\angle B = \angle B$ (Common angle)
Therefore,by $AA$ similarity criterion,$\triangle BCA \sim \triangle BAD$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{BC}{BA} = \frac{BA}{BD}$
Cross-multiplying gives:
$BA \cdot BA = BC \cdot BD$
$AB^2 = BC \cdot BD$.
Hence proved.

(N/A) Given: In $\triangle ABD$,$\angle A = 90^{\circ}$ and $AC \perp BD$.
To prove: $AC^{2} = BC \cdot DC$.
Proof:
In $\triangle BCA$ and $\triangle ACD$:
$\angle BCA = \angle ACD = 90^{\circ}$ (Since $AC \perp BD$)
$\angle CBA = 90^{\circ} - \angle CAB$ (In $\triangle ABC$,$\angle B + \angle CAB = 90^{\circ}$)
Also,$\angle CAD = 90^{\circ} - \angle CAB$ (Since $\angle A = 90^{\circ}$,$\angle CAD + \angle CAB = 90^{\circ}$)
Therefore,$\angle CBA = \angle CAD$.
By $AA$ similarity criterion,$\triangle BCA \sim \triangle ACD$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{BC}{AC} = \frac{AC}{DC}$
Cross-multiplying,we get:
$AC^{2} = BC \cdot DC$.
Hence proved.

(N/A) Given: In $\triangle ABD$,$\angle A = 90^{\circ}$ and $AC \perp BD$.
To prove: $AD^{2} = BD \cdot CD$.
Proof: Consider $\triangle ACD$ and $\triangle BCA$.
In $\triangle ACD$ and $\triangle BCA$:
$\angle ACD = \angle BCA = 90^{\circ}$ (Since $AC \perp BD$ and $\angle A = 90^{\circ}$).
$\angle CAD = \angle CBA$ (Since both are complementary to $\angle BAD$ or $\angle BDA$).
Therefore,by $AA$ similarity criterion,$\triangle ACD \sim \triangle BCA$.
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{CD}{AC} = \frac{AC}{BC} = \frac{AD}{AB}$.
Wait,let us consider $\triangle ACD$ and $\triangle BAD$.
In $\triangle ACD$ and $\triangle BAD$:
$\angle ACD = \angle BAD = 90^{\circ}$.
$\angle D = \angle D$ (Common angle).
Therefore,by $AA$ similarity criterion,$\triangle ACD \sim \triangle BAD$.
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{AC}{BA} = \frac{CD}{AD} = \frac{AD}{BD}$.
Taking the last two parts of the ratio:
$\frac{CD}{AD} = \frac{AD}{BD}$.
Cross-multiplying gives:
$AD^{2} = BD \cdot CD$.
Hence proved.

(N/A) Given that $\triangle ABC$ is an isosceles triangle with $\angle C = 90^{\circ}$.
Since it is an isosceles triangle,the two legs must be equal,so $AC = CB$.
Applying the Pythagoras theorem in $\triangle ABC$ (which is right-angled at point $C$),we have:
$AC^{2} + CB^{2} = AB^{2}$
Since $AC = CB$,we can substitute $CB$ with $AC$ in the equation:
$AC^{2} + AC^{2} = AB^{2}$
$2AC^{2} = AB^{2}$
Hence,it is proved that $AB^{2} = 2AC^{2}$.

(N/A) Given that,
$AB^2 = 2 AC^2$
We can rewrite this as:
$AB^2 = AC^2 + AC^2$
Since it is given that $AC = BC$,we can substitute $BC$ for one of the $AC$ terms:
$AB^2 = AC^2 + BC^2$
This equation satisfies the converse of the Pythagoras theorem,which states that if the square of one side of a triangle is equal to the sum of the squares of the other two sides,then the angle opposite the first side is a right angle.
Therefore,$\triangle ABC$ is a right-angled triangle with the right angle at $C$.

(N/A) Let $AD$ be the altitude in the given equilateral triangle,$\triangle ABC$.
We know that an altitude in an equilateral triangle bisects the opposite side.
$\therefore BD = DC = a$
In $\triangle ADB$,we have $\angle ADB = 90^{\circ}$.
Applying the Pythagoras theorem,we obtain:
$AD^2 + DB^2 = AB^2$
$\Rightarrow AD^2 + a^2 = (2a)^2$
$\Rightarrow AD^2 + a^2 = 4a^2$
$\Rightarrow AD^2 = 3a^2$
$\Rightarrow AD = a\sqrt{3}$
In an equilateral triangle,all the altitudes are equal in length.
Therefore,the length of each altitude is $a\sqrt{3}$.

(N/A) Let $ABCD$ be a rhombus with diagonals $AC$ and $BD$ intersecting at $O$.
In $\Delta AOB, \Delta BOC, \Delta COD,$ and $\Delta AOD,$
Applying the Pythagoras theorem,we obtain:
$AB^2 = AO^2 + OB^2$ $...(1)$
$BC^2 = BO^2 + OC^2$ $...(2)$
$CD^2 = CO^2 + OD^2$ $...(3)$
$AD^2 = AO^2 + OD^2$ $...(4)$
Adding all these equations,we obtain:
$AB^2 + BC^2 + CD^2 + AD^2 = 2(AO^2 + OB^2 + OC^2 + OD^2)$
Since the diagonals of a rhombus bisect each other at right angles,$AO = OC = AC/2$ and $BO = OD = BD/2$.
Substituting these values:
$= 2((AC/2)^2 + (BD/2)^2 + (AC/2)^2 + (BD/2)^2)$
$= 2(2(AC/2)^2 + 2(BD/2)^2)$
$= 2(AC^2/2 + BD^2/2)$
$= AC^2 + BD^2$
Thus,the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

(N/A) Join $OA$,$OB$,and $OC$.
Applying the Pythagoras theorem in $\triangle AOF$,we obtain:
$OA^{2} = OF^{2} + AF^{2}$ --- $(i)$
Similarly,in $\triangle BOD$:
$OB^{2} = OD^{2} + BD^{2}$ --- $(ii)$
Similarly,in $\triangle COE$:
$OC^{2} = OE^{2} + CE^{2}$ --- $(iii)$
Adding equations $(i)$,$(ii)$,and $(iii)$,we get:
$OA^{2} + OB^{2} + OC^{2} = OF^{2} + AF^{2} + OD^{2} + BD^{2} + OE^{2} + CE^{2}$
Rearranging the terms,we get:
$OA^{2} + OB^{2} + OC^{2} - OD^{2} - OE^{2} - OF^{2} = AF^{2} + BD^{2} + CE^{2}$
Hence,the result is proved.

(N/A) Join $OA$,$OB$,and $OC$.
In right-angled triangles $\triangle AFO$,$\triangle BDO$,and $\triangle CEO$,by applying the Pythagoras theorem:
$OA^{2} = AF^{2} + OF^{2} \implies AF^{2} = OA^{2} - OF^{2}$ ... $(i)$
$OB^{2} = BD^{2} + OD^{2} \implies BD^{2} = OB^{2} - OD^{2}$ ... $(ii)$
$OC^{2} = CE^{2} + OE^{2} \implies CE^{2} = OC^{2} - OE^{2}$ ... $(iii)$
Adding $(i)$,$(ii)$,and $(iii)$,we get:
$AF^{2} + BD^{2} + CE^{2} = (OA^{2} - OF^{2}) + (OB^{2} - OD^{2}) + (OC^{2} - OE^{2})$ ... $(iv)$
Similarly,in right-angled triangles $\triangle AEO$,$\triangle CDO$,and $\triangle BFO$:
$OA^{2} = AE^{2} + OE^{2} \implies AE^{2} = OA^{2} - OE^{2}$
$OC^{2} = CD^{2} + OD^{2} \implies CD^{2} = OC^{2} - OD^{2}$
$OB^{2} = BF^{2} + OF^{2} \implies BF^{2} = OB^{2} - OF^{2}$
Adding these,we get:
$AE^{2} + CD^{2} + BF^{2} = (OA^{2} - OE^{2}) + (OC^{2} - OD^{2}) + (OB^{2} - OF^{2})$
Rearranging the terms,this is equal to the right-hand side of equation $(iv)$.
Thus,$AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}$.

(A) Let $OA$ be the wall and $AB$ be the ladder,where $OA = 8\, m$ and $AB = 10\, m$.
Since the wall is perpendicular to the ground,$\triangle AOB$ is a right-angled triangle with $\angle AOB = 90^\circ$.
By Pythagoras theorem,$AB^2 = OA^2 + OB^2$.
Substituting the values: $(10)^2 = (8)^2 + OB^2$.
$100 = 64 + OB^2$.
$OB^2 = 100 - 64 = 36$.
$OB = \sqrt{36} = 6\, m$.
Therefore,the distance of the foot of the ladder from the base of the wall is $6\, m$.

(N/A) Let $OB$ be the pole and $AB$ be the wire. The pole is vertical,so $\triangle AOB$ is a right-angled triangle with $\angle AOB = 90^{\circ}$.
By Pythagoras theorem,
$AB^{2} = OB^{2} + OA^{2}$
$(24\, m)^{2} = (18\, m)^{2} + OA^{2}$
$576\, m^{2} = 324\, m^{2} + OA^{2}$
$OA^{2} = (576 - 324)\, m^{2} = 252\, m^{2}$
$OA = \sqrt{252}\, m = \sqrt{36 \times 7}\, m = 6\sqrt{7}\, m$
Therefore,the distance from the base is $6\sqrt{7}\, m$ (approximately $15.87\, m$).

(N/A) Distance travelled by the plane flying towards north in $1 \frac{1}{2}$ hours $= 1000 \times 1.5 = 1500\, km$.
Distance travelled by the plane flying towards west in $1 \frac{1}{2}$ hours $= 1200 \times 1.5 = 1800\, km$.
Let the airport be at the origin $O$. The plane flying north reaches point $A$ and the plane flying west reaches point $B$.
Since north and west directions are perpendicular,$\triangle AOB$ is a right-angled triangle with $\angle AOB = 90^\circ$.
By Pythagoras theorem,the distance between the two planes is the hypotenuse $AB = \sqrt{OA^2 + OB^2}$.
$AB = \sqrt{1500^2 + 1800^2} = \sqrt{2250000 + 3240000} = \sqrt{5490000}$.
$AB = \sqrt{90000 \times 61} = 300\sqrt{61}\, km$.
Thus,the distance between the two planes after $1 \frac{1}{2}$ hours is $300\sqrt{61}\, km$.

(13 M) Let $CD$ and $AB$ be the poles of height $11 \, m$ and $6 \, m$ respectively.
Draw a line segment $AP$ parallel to $BD$ such that $P$ lies on $CD$. Then $AP = BD = 12 \, m$ and $PD = AB = 6 \, m$.
Now,$CP = CD - PD = 11 \, m - 6 \, m = 5 \, m$.
In the right-angled triangle $\triangle APC$,by applying the Pythagoras theorem:
$AC^2 = AP^2 + CP^2$
$AC^2 = (12 \, m)^2 + (5 \, m)^2$
$AC^2 = 144 \, m^2 + 25 \, m^2 = 169 \, m^2$
$AC = \sqrt{169} \, m = 13 \, m$.
Therefore,the distance between their tops is $13 \, m$.

(N/A) Applying the Pythagoras theorem in $\triangle ACE$,we obtain:
$AC^{2} + CE^{2} = AE^{2}$ $...(1)$
Applying the Pythagoras theorem in $\triangle BCD$,we obtain:
$BC^{2} + CD^{2} = BD^{2}$ $...(2)$
Adding equation $(1)$ and equation $(2)$,we obtain:
$AC^{2} + CE^{2} + BC^{2} + CD^{2} = AE^{2} + BD^{2}$ $...(3)$
Applying the Pythagoras theorem in $\triangle CDE$,we obtain:
$DE^{2} = CD^{2} + CE^{2}$ $...(4)$
Applying the Pythagoras theorem in $\triangle ABC$,we obtain:
$AB^{2} = AC^{2} + BC^{2}$ $...(5)$
Substituting the values from equation $(4)$ and equation $(5)$ into equation $(3)$,we obtain:
$AB^{2} + DE^{2} = AE^{2} + BD^{2}$
Hence,$AE^{2} + BD^{2} = AB^{2} + DE^{2}$.

(N/A) Applying the Pythagoras theorem for $\triangle ACD$,we obtain:
$AC^{2} = AD^{2} + DC^{2}$
$AD^{2} = AC^{2} - DC^{2}$ $...(1)$
Applying the Pythagoras theorem for $\triangle ABD$,we obtain:
$AB^{2} = AD^{2} + DB^{2}$
$AD^{2} = AB^{2} - DB^{2}$ $...(2)$
From equation $(1)$ and equation $(2)$,we obtain:
$AC^{2} - DC^{2} = AB^{2} - DB^{2}$ $...(3)$
It is given that $DB = 3CD$. Since $BC = DB + CD$,we have $BC = 3CD + CD = 4CD$.
$\therefore CD = \frac{BC}{4}$ and $DB = 3CD = \frac{3BC}{4}$.
Substituting these values into equation $(3)$,we obtain:
$AC^{2} - \left(\frac{BC}{4}\right)^{2} = AB^{2} - \left(\frac{3BC}{4}\right)^{2}$
$AC^{2} - \frac{BC^{2}}{16} = AB^{2} - \frac{9BC^{2}}{16}$
$16AC^{2} - BC^{2} = 16AB^{2} - 9BC^{2}$
$16AB^{2} - 16AC^{2} = 8BC^{2}$
Dividing by $8$,we get:
$2AB^{2} = 2AC^{2} + BC^{2}$

(N/A) Let the side of the equilateral triangle be $a$,and $AE$ be the altitude of $\Delta ABC$ drawn to $BC$.
Since $AE$ is the altitude of an equilateral triangle,it bisects the base $BC$.
$\therefore BE = EC = \frac{BC}{2} = \frac{a}{2}$.
In the right-angled triangle $\Delta ABE$,by Pythagoras theorem:
$AE^{2} = AB^{2} - BE^{2} = a^{2} - (\frac{a}{2})^{2} = a^{2} - \frac{a^{2}}{4} = \frac{3a^{2}}{4}$.
So,$AE = \frac{a\sqrt{3}}{2}$.
Given that $BD = \frac{1}{3} BC = \frac{a}{3}$.
Now,$DE = BE - BD = \frac{a}{2} - \frac{a}{3} = \frac{3a - 2a}{6} = \frac{a}{6}$.
In the right-angled triangle $\Delta ADE$,by Pythagoras theorem:
$AD^{2} = AE^{2} + DE^{2}$.
Substituting the values:
$AD^{2} = (\frac{a\sqrt{3}}{2})^{2} + (\frac{a}{6})^{2} = \frac{3a^{2}}{4} + \frac{a^{2}}{36}$.
Taking $LCM$ as $36$:
$AD^{2} = \frac{27a^{2} + a^{2}}{36} = \frac{28a^{2}}{36} = \frac{7a^{2}}{9}$.
Since $a = AB$,we have $AD^{2} = \frac{7 AB^{2}}{9}$.
Therefore,$9 AD^{2} = 7 AB^{2}$.
Hence proved.

(N/A) Let the side of the equilateral triangle be $a$,and $AE$ be the altitude of $\triangle ABC$.
Since the altitude of an equilateral triangle bisects the base,
$BE = EC = \frac{BC}{2} = \frac{a}{2}$
Applying the Pythagoras theorem in the right-angled $\triangle ABE$,we obtain:
$AB^2 = AE^2 + BE^2$
Substituting the values $AB = a$ and $BE = \frac{a}{2}$:
$a^2 = AE^2 + (\frac{a}{2})^2$
$a^2 = AE^2 + \frac{a^2}{4}$
Rearranging the terms to solve for $AE^2$:
$AE^2 = a^2 - \frac{a^2}{4}$
$AE^2 = \frac{4a^2 - a^2}{4}$
$AE^2 = \frac{3a^2}{4}$
Multiplying both sides by $4$:
$4AE^2 = 3a^2$
Thus,it is proved that $4 \times$ (square of altitude) $= 3 \times$ (square of one side).

(B) Given that,$AB = 6\sqrt{3} \text{ cm}$,$AC = 12 \text{ cm}$,and $BC = 6 \text{ cm}$.
Calculating the squares of the sides:
$AB^2 = (6\sqrt{3})^2 = 36 \times 3 = 108$
$BC^2 = (6)^2 = 36$
$AC^2 = (12)^2 = 144$
Now,observe that:
$AB^2 + BC^2 = 108 + 36 = 144$
Since $AB^2 + BC^2 = AC^2$,the triangle satisfies the converse of the Pythagoras theorem.
Therefore,$\triangle ABC$ is a right-angled triangle,with the right angle opposite to the hypotenuse $AC$.
Thus,$\angle B = 90^\circ$.
Hence,the correct answer is $(B)$.

(N/A) Construction: Draw a line segment $RT$ parallel to $PS$ which intersects the extended line segment $QP$ at point $T$.
Given that,$PS$ is the angle bisector of $\angle QPR$.
Therefore,$\angle QPS = \angle SPR \quad \dots(1)$
By construction,$PS \parallel TR$ and $QT$ is a transversal.
Therefore,$\angle QPS = \angle QTR$ (Corresponding angles) $\quad \dots(2)$
Also,$PS \parallel TR$ and $PR$ is a transversal.
Therefore,$\angle SPR = \angle PRT$ (Alternate interior angles) $\quad \dots(3)$
From equations $(1)$,$(2)$,and $(3)$,we get:
$\angle QTR = \angle PRT$
In $\Delta PTR$,since $\angle QTR = \angle PRT$,the sides opposite to these angles are equal.
Therefore,$PT = PR \quad \dots(4)$
In $\Delta QTR$,since $PS \parallel TR$,by the Basic Proportionality Theorem $(BPT)$:
$\frac{QS}{SR} = \frac{QP}{PT}$
Substituting $PT = PR$ from equation $(4)$ into the above expression:
$\frac{QS}{SR} = \frac{PQ}{PR}$
Hence,it is proved.

(N/A) Join $DB.$
We have $DN \parallel CB,$ $DM \parallel AB,$ and $\angle B = 90^{\circ}.$
Therefore,$DMBN$ is a rectangle.
Thus,$DN = MB$ and $DM = NB.$
Since $D$ is the foot of the perpendicular drawn from $B$ to $AC,$ we have $\angle CDB = 90^{\circ}.$
$\angle 2 + \angle 3 = 90^{\circ} \dots(1)$
In $\Delta CDM, \angle 1 + \angle 2 + \angle DMC = 180^{\circ} \Rightarrow \angle 1 + \angle 2 = 90^{\circ} \dots(2)$
In $\Delta DMB, \angle 3 + \angle DMB + \angle 4 = 180^{\circ} \Rightarrow \angle 3 + \angle 4 = 90^{\circ} \dots(3)$
From $(1)$ and $(2),$ we get $\angle 1 = \angle 3.$
From $(1)$ and $(3),$ we get $\angle 2 = \angle 4.$
In $\Delta DCM$ and $\Delta BDM,$
$\angle 1 = \angle 3$ (Proved above)
$\angle 2 = \angle 4$ (Proved above)
Therefore,$\Delta DCM \sim \Delta BDM$ ($AA$ similarity criterion).
Thus,$\frac{BM}{DM} = \frac{DM}{MC}.$
Since $BM = DN,$ we have $\frac{DN}{DM} = \frac{DM}{MC}.$
Therefore,$DM^{2} = DN \cdot MC.$

(N/A) In quadrilateral $DMBN,$ since $\angle DMB = 90^{\circ},$ $\angle DNB = 90^{\circ},$ and $\angle MBN = 90^{\circ}$ (given $\angle ABC = 90^{\circ}$),the fourth angle $\angle MDN$ must also be $90^{\circ}.$ Thus,$DMBN$ is a rectangle.
Therefore,$DM = NB$ and $DN = MB.$
In $\Delta ABC,$ $BD \perp AC.$ By the property of similar triangles in a right triangle,$\Delta ADN \sim \Delta ABD$ and $\Delta BDC \sim \Delta ABD.$
More directly,in $\Delta DNB$ and $\Delta ADN:$
$\angle DNB = \angle DNA = 90^{\circ}.$
$\angle ND B = \angle DAN$ (since both are complementary to $\angle ADN$).
Thus,$\Delta DNB \sim \Delta ADN$ by $AA$ similarity.
Therefore,$\frac{DN}{AN} = \frac{NB}{DN}.$
$DN^{2} = AN \cdot NB.$
Since $NB = DM,$ we have $DN^{2} = AN \cdot DM.$

(N/A) In $\triangle ADB$,applying the Pythagoras theorem,we have:
$AB^{2} = AD^{2} + BD^{2} \quad \dots(1)$
In $\triangle ADC$,applying the Pythagoras theorem,we have:
$AC^{2} = AD^{2} + DC^{2}$
Since $DC = DB + BC$,we can write:
$AC^{2} = AD^{2} + (DB + BC)^{2}$
$AC^{2} = AD^{2} + DB^{2} + BC^{2} + 2DB \cdot BC$
Substituting $AD^{2} + DB^{2} = AB^{2}$ from equation $(1)$:
$AC^{2} = AB^{2} + BC^{2} + 2BC \cdot BD$

(N/A) Applying the Pythagoras theorem in $\triangle ADB$,we obtain:
$AD^{2} + DB^{2} = AB^{2}$
$\Rightarrow AD^{2} = AB^{2} - DB^{2} \dots(1)$
Applying the Pythagoras theorem in $\triangle ADC$,we obtain:
$AD^{2} + DC^{2} = AC^{2}$
Substituting the value of $AD^{2}$ from equation $(1)$:
$(AB^{2} - BD^{2}) + DC^{2} = AC^{2}$
Since $D$ lies on $BC$,we have $DC = BC - BD$. Substituting this into the equation:
$AB^{2} - BD^{2} + (BC - BD)^{2} = AC^{2}$
$AB^{2} - BD^{2} + (BC^{2} + BD^{2} - 2BC \cdot BD) = AC^{2}$
$AB^{2} + BC^{2} - 2BC \cdot BD = AC^{2}$
Thus,$AC^{2} = AB^{2} + BC^{2} - 2BC \cdot BD$.

(N/A) Applying Pythagoras theorem in $\Delta AMD$,we obtain:
$AM^{2} + MD^{2} = AD^{2} \dots(1)$
Applying Pythagoras theorem in $\Delta AMC$,we obtain:
$AM^{2} + MC^{2} = AC^{2}$
Since $MC = MD + DC$,we have:
$AM^{2} + (MD + DC)^{2} = AC^{2}$
$AM^{2} + MD^{2} + DC^{2} + 2MD \cdot DC = AC^{2}$
Substituting $AM^{2} + MD^{2} = AD^{2}$ from equation $(1)$:
$AD^{2} + DC^{2} + 2MD \cdot DC = AC^{2}$
Since $AD$ is a median,$DC = \frac{BC}{2}$. Substituting this value:
$AD^{2} + \left(\frac{BC}{2}\right)^{2} + 2MD \cdot \left(\frac{BC}{2}\right) = AC^{2}$
$AD^{2} + \left(\frac{BC}{2}\right)^{2} + BC \cdot DM = AC^{2}$
Thus,$AC^{2} = AD^{2} + BC \cdot DM + \left(\frac{BC}{2}\right)^{2}$.

(N/A) Given: $AD$ is the median of $\Delta ABC$,so $BD = DC = \frac{BC}{2}$.
Also,$AM \perp BC$.
In right-angled $\Delta ABM$,by Pythagoras theorem:
$AB^{2} = AM^{2} + BM^{2}$
In right-angled $\Delta ADM$,by Pythagoras theorem:
$AD^{2} = AM^{2} + DM^{2} \implies AM^{2} = AD^{2} - DM^{2}$
Substituting $AM^{2}$ in the first equation:
$AB^{2} = (AD^{2} - DM^{2}) + BM^{2}$
Since $BM = BD - DM$,we have:
$AB^{2} = AD^{2} - DM^{2} + (BD - DM)^{2}$
$AB^{2} = AD^{2} - DM^{2} + BD^{2} + DM^{2} - 2 \cdot BD \cdot DM$
$AB^{2} = AD^{2} + BD^{2} - 2 \cdot BD \cdot DM$
Since $BD = \frac{BC}{2}$,substituting this value:
$AB^{2} = AD^{2} + \left(\frac{BC}{2}\right)^{2} - 2 \cdot \left(\frac{BC}{2}\right) \cdot DM$
$AB^{2} = AD^{2} + \left(\frac{BC}{2}\right)^{2} - BC \cdot DM$

(N/A) Applying Pythagoras theorem in $\Delta ABM$,we obtain:
$AM^2 + MB^2 = AB^2 \quad \dots(1)$
Applying Pythagoras theorem in $\Delta AMC$,we obtain:
$AM^2 + MC^2 = AC^2 \quad \dots(2)$
Adding equations $(1)$ and $(2)$,we obtain:
$2AM^2 + MB^2 + MC^2 = AB^2 + AC^2$
Since $AD$ is a median,$BD = DC = \frac{BC}{2}$.
We can write $MB = BD - DM$ and $MC = DC + DM = BD + DM$.
Substituting these in the equation:
$2AM^2 + (BD - DM)^2 + (BD + DM)^2 = AB^2 + AC^2$
$2AM^2 + (BD^2 + DM^2 - 2BD \cdot DM) + (BD^2 + DM^2 + 2BD \cdot DM) = AB^2 + AC^2$
$2AM^2 + 2BD^2 + 2DM^2 = AB^2 + AC^2$
$2(AM^2 + DM^2) + 2BD^2 = AB^2 + AC^2$
In $\Delta ADM$,by Pythagoras theorem,$AM^2 + DM^2 = AD^2$.
So,$2AD^2 + 2(\frac{BC}{2})^2 = AB^2 + AC^2$
$2AD^2 + 2(\frac{BC^2}{4}) = AB^2 + AC^2$
$2AD^2 + \frac{BC^2}{2} = AB^2 + AC^2$

(A) Let $ABCD$ be a parallelogram.
Let us draw perpendicular $DE$ on the extended side $AB$,and $AF$ on side $DC$.
Applying Pythagoras theorem in $\triangle DEA$,we obtain:
$DE^{2} + EA^{2} = DA^{2} \dots (i)$
Applying Pythagoras theorem in $\triangle DEB$,we obtain:
$DE^{2} + EB^{2} = DB^{2}$
$DE^{2} + (EA + AB)^{2} = DB^{2}$
$(DE^{2} + EA^{2}) + AB^{2} + 2EA \times AB = DB^{2}$
$DA^{2} + AB^{2} + 2EA \times AB = DB^{2} \dots (ii)$
Applying Pythagoras theorem in $\triangle AFC$,we obtain:
$AC^{2} = AF^{2} + FC^{2}$
$AC^{2} = AF^{2} + (DC - FD)^{2}$
$AC^{2} = AF^{2} + DC^{2} + FD^{2} - 2DC \times FD$
$AC^{2} = (AF^{2} + FD^{2}) + DC^{2} - 2DC \times FD$
$AC^{2} = AD^{2} + DC^{2} - 2DC \times FD \dots (iii)$
Since $ABCD$ is a parallelogram,$AB = CD \dots (iv)$ and $BC = AD \dots (v)$.
In $\triangle DEA$ and $\triangle AFD$:
$\angle DEA = \angle AFD = 90^{\circ}$
$\angle EAD = \angle ADF$ (since $EA \parallel DF$)
$AD = AD$ (Common side)
Therefore,$\triangle EAD \cong \triangle FDA$ ($AAS$ congruence criterion).
This implies $EA = DF \dots (vi)$.
Adding equations $(ii)$ and $(iii)$:
$DB^{2} + AC^{2} = DA^{2} + AB^{2} + 2EA \times AB + AD^{2} + DC^{2} - 2DC \times FD$
Using $AB = DC$ and $EA = DF$:
$DB^{2} + AC^{2} = AD^{2} + AB^{2} + 2EA \times AB + AD^{2} + AB^{2} - 2AB \times EA$
$DB^{2} + AC^{2} = 2AD^{2} + 2AB^{2}$
Since $AD = BC$ and $AB = CD$:
$AC^{2} + BD^{2} = AB^{2} + BC^{2} + CD^{2} + DA^{2}$.

(N/A) To prove that $\Delta APC \sim \Delta DPB$,we follow these steps:
$1$. Join $CB$ and $AD$.
$2$. Consider $\Delta APC$ and $\Delta DPB$.
$3$. $\angle APC = \angle DPB$ (Vertically opposite angles).
$4$. $\angle CAP = \angle BDP$ (Angles subtended by the same arc $CB$ in the same segment of the circle are equal).
$5$. Therefore,by the $AA$ (Angle-Angle) similarity criterion,$\Delta APC \sim \Delta DPB$.

(N/A) To prove $AP \cdot PB = CP \cdot DP$,we consider the triangles $\Delta APC$ and $\Delta DPB$.
$1$. $\angle APC = \angle DPB$ (Vertically opposite angles).
$2$. $\angle CAP = \angle BDP$ (Angles in the same segment of a circle are equal).
By the $AA$ (Angle-Angle) similarity criterion,$\Delta APC \sim \Delta DPB$.
Since the corresponding sides of similar triangles are proportional,we have:
$\frac{AP}{DP} = \frac{PC}{PB} = \frac{AC}{DB}$
Taking the first two ratios:
$\frac{AP}{DP} = \frac{PC}{PB}$
By cross-multiplying,we get:
$AP \cdot PB = PC \cdot DP$ (or $AP \cdot PB = CP \cdot DP$).

(N/A) Consider $\Delta PAC$ and $\Delta PDB$.
$1$. $\angle P = \angle P$ (Common angle).
$2$. In a cyclic quadrilateral $ABDC$,the exterior angle is equal to the interior opposite angle. Therefore,$\angle PAC = \angle PDB$ (Since $\angle PAC + \angle CAB = 180^{\circ}$ and $\angle PDB + \angle CAB = 180^{\circ}$).
$3$. By the $AA$ (Angle-Angle) similarity criterion,$\Delta PAC \sim \Delta PDB$.

(N/A) To prove $PA \cdot PB = PC \cdot PD$,consider $\triangle PAD$ and $\triangle PCB$.
$1$. $\angle P = \angle P$ (Common angle).
$2$. $\angle PDA = \angle PBC$ (Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle,or since $\angle PAD + \angle DAB = 180^{\circ}$ and $\angle BCD + \angle DAB = 180^{\circ}$,we have $\angle PAD = \angle BCD$,and thus $\triangle PAD \sim \triangle PCB$ by $AA$ similarity).
Since $\triangle PAD \sim \triangle PCB$,the ratios of their corresponding sides are equal:
$\frac{PA}{PC} = \frac{PD}{PB} = \frac{AD}{CB}$.
Taking the first two parts:
$\frac{PA}{PC} = \frac{PD}{PB}$
By cross-multiplying,we get:
$PA \cdot PB = PC \cdot PD$.

(N/A) Let us extend $BA$ to $P$ such that $AP = AC$. Join $PC$.
It is given that,
$\frac{BD}{CD} = \frac{AB}{AC}$
Since $AP = AC$,we can substitute $AP$ for $AC$:
$\frac{BD}{CD} = \frac{AB}{AP}$
By using the converse of the Basic Proportionality Theorem $(BPT)$ in $\Delta BPC$,we obtain:
$AD \parallel PC$
Therefore,$\angle BAD = \angle APC$ (Corresponding angles) $\dots(1)$
And,$\angle DAC = \angle ACP$ (Alternate interior angles) $\dots(2)$
By construction,we have $AP = AC$,which implies that in $\Delta APC$,the angles opposite to equal sides are equal:
$\angle APC = \angle ACP \dots(3)$
From equations $(1), (2),$ and $(3)$,we get:
$\angle BAD = \angle DAC$
Thus,$AD$ is the bisector of $\angle BAC$.