Mix Examples - Triangles Questions in English

(N/A) Given: $A$ square $ABCD$ and $\triangle OAB$ is an equilateral triangle,so $OA = OB = AB$.
To prove: $\triangle OCD$ is an isosceles triangle.
Proof: In square $ABCD$,$\angle DAB = \angle CBA = 90^{\circ}$.
Since $\triangle OAB$ is equilateral,$\angle OAB = \angle OBA = 60^{\circ}$.
Now,$\angle DAO = \angle DAB - \angle OAB = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Similarly,$\angle CBO = \angle CBA - \angle OBA = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
In $\triangle DAO$ and $\triangle CBO$:
$AD = BC$ (Sides of a square)
$\angle DAO = \angle CBO = 30^{\circ}$ (Proved above)
$OA = OB$ (Sides of an equilateral triangle)
By $SAS$ congruence criterion,$\triangle DAO \cong \triangle CBO$.
Therefore,$OD = OC$ (Corresponding parts of congruent triangles).
Since $OD = OC$,$\triangle OCD$ is an isosceles triangle.
Hence,proved.

(N/A) Given: $\triangle ABC$ and $\triangle DBC$ on the same base $BC$. Also,$AB = AC$ and $BD = CD$.
To prove: $AD$ is the perpendicular bisector of $BC$,i.e.,$OB = OC$ and $\angle AOB = \angle AOC = 90^{\circ}$.
Proof: In $\triangle ABD$ and $\triangle ACD$,we have:
$AB = AC$ [Given]
$BD = CD$ [Given]
$AD = AD$ [Common side]
So,by $SSS$ criterion of congruence,we have:
$\triangle ABD \cong \triangle ACD$
Therefore,$\angle 1 = \angle 2$ [$CPCT$]
Now,in $\triangle ABO$ and $\triangle ACO$,we have:
$AB = AC$ [Given]
$\angle 1 = \angle 2$ [Proved above]
$AO = AO$ [Common side]
So,by $SAS$ criterion of congruence,we have:
$\triangle ABO \cong \triangle ACO$
Therefore,$BO = CO$ [$CPCT$]
And,$\angle 3 = \angle 4$ [$CPCT$]
But,$\angle 3 + \angle 4 = 180^{\circ}$ [Linear pair axiom]
$\Rightarrow \angle 3 + \angle 3 = 180^{\circ}$
$\Rightarrow 2\angle 3 = 180^{\circ}$
$\Rightarrow \angle 3 = 90^{\circ}$
Since $BO = CO$ and $\angle 3 = 90^{\circ}$,$AD$ is the perpendicular bisector of $BC$. Hence,proved.

(N/A) In $\triangle ADC$ and $\triangle BEC$ we have:
$AC = BC$ [Given] ... $(1)$
$\angle ADC = \angle BEC = 90^{\circ}$ [Given as altitudes]
$\angle ACD = \angle BCE$ [Common angle]
Therefore,$\triangle ADC \cong \triangle BEC$ [By $AAS$ congruence rule]
Therefore,$CD = CE$ ... $(2)$ [$CPCT$]
Subtracting $(2)$ from $(1)$,we get:
$AC - CE = BC - CD$
$AE = BD$
Hence,proved.

(N/A) Given: $\triangle ABC$ with median $AD$ to side $BC$.
To prove: $AB + AC > 2AD$.
Construction: Produce $AD$ to $E$ such that $DE = AD$ and join $EC$.
Proof: In $\triangle ADB$ and $\triangle EDC$:
$AD = ED$ (By construction)
$\angle 1 = \angle 2$ (Vertically opposite angles)
$DB = DC$ ($AD$ is the median)
By $SAS$ congruence criterion,$\triangle ADB \cong \triangle EDC$.
Therefore,$AB = EC$ $(CPCT)$.
Now,in $\triangle AEC$,the sum of any two sides is greater than the third side:
$AC + CE > AE$
Since $AE = AD + DE = AD + AD = 2AD$ and $CE = AB$:
$AC + AB > 2AD$.
Hence,it is proved that the sum of two sides is greater than twice the median to the third side.

(N/A) Given: $A$ quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $O$.
To prove: $AB + BC + CD + DA < 2(BD + AC)$.
Proof: In $\triangle AOB$,by the triangle inequality theorem (the sum of any two sides of a triangle is greater than the third side),we have:
$OA + OB > AB$ ... $(1)$
In $\triangle BOC$,we have:
$OB + OC > BC$ ... $(2)$
In $\triangle COD$,we have:
$OC + OD > CD$ ... $(3)$
In $\triangle DOA$,we have:
$OD + OA > DA$ ... $(4)$
Adding equations $(1), (2), (3),$ and $(4)$,we get:
$(OA + OB) + (OB + OC) + (OC + OD) + (OD + OA) > AB + BC + CD + DA$
$2(OA + OC) + 2(OB + OD) > AB + BC + CD + DA$
Since $OA + OC = AC$ and $OB + OD = BD$,we substitute these into the inequality:
$2(AC) + 2(BD) > AB + BC + CD + DA$
$2(AC + BD) > AB + BC + CD + DA$
Therefore,$AB + BC + CD + DA < 2(AC + BD)$.
Hence,proved.

(N/A) Given: $A$ quadrilateral $ABCD$.
To prove: $AB + BC + CD + DA > AC + BD$.
Proof: In $\triangle ABC$,we have $AB + BC > AC$ $\ldots(1)$ [Since the sum of the lengths of any two sides of a triangle must be greater than the third side].
In $\triangle BCD$,we have $BC + CD > BD$ $\ldots(2)$ [Same reason].
In $\triangle CDA$,we have $CD + DA > AC$ $\ldots(3)$ [Same reason].
In $\triangle DAB$,we have $DA + AB > BD$ $\ldots(4)$ [Same reason].
Adding $(1), (2), (3)$ and $(4)$,we get:
$(AB + BC) + (BC + CD) + (CD + DA) + (DA + AB) > AC + BD + AC + BD$
$2AB + 2BC + 2CD + 2DA > 2AC + 2BD$
$2(AB + BC + CD + DA) > 2(AC + BD)$
$AB + BC + CD + DA > AC + BD$.
Hence,proved.

(N/A) We have to prove that $\angle ABC = 90^{\circ}$.
Since $D$ is the mid-point of $AC$,we have $AD = DC$.
Also,it is given that $BD = \frac{1}{2} AC$.
Since $AC = AD + DC = 2AD$,we have $BD = AD$.
Therefore,$BD = AD = DC$.
In $\Delta ABD$,since $BD = AD$,the angles opposite to equal sides are equal,so $\angle 1 = \angle 2$. $(1)$
In $\Delta BCD$,since $BD = DC$,the angles opposite to equal sides are equal,so $\angle 3 = \angle 4$. $(2)$
In $\Delta ABC$,the sum of angles is $180^{\circ}$:
$\angle A + \angle ABC + \angle C = 180^{\circ}$
$\angle 1 + (\angle 2 + \angle 3) + \angle 4 = 180^{\circ}$
Substituting $\angle 1 = \angle 2$ and $\angle 4 = \angle 3$ from $(1)$ and $(2)$:
$\angle 2 + (\angle 2 + \angle 3) + \angle 3 = 180^{\circ}$
$2(\angle 2 + \angle 3) = 180^{\circ}$
$\angle 2 + \angle 3 = 90^{\circ}$
Since $\angle ABC = \angle 2 + \angle 3$,we have $\angle ABC = 90^{\circ}$.
Hence proved.

(N/A) $ABC$ is a right triangle,right-angled at $B$,and $D$ is the mid-point of the hypotenuse $AC$. We need to prove that $BD = \frac{1}{2} AC$.
Construction: Produce $BD$ to $E$ such that $BD = DE$. Join $EC$.
In $\Delta ADB$ and $\Delta CDE$:
$AD = CD$ (Since $D$ is the mid-point of $AC$)
$\angle ADB = \angle CDE$ (Vertically opposite angles)
$BD = DE$ (By construction)
Therefore,$\Delta ADB \cong \Delta CDE$ (By $SAS$ congruence criterion).
Thus,$AB = EC$ $(CPCT)$ and $\angle 1 = \angle 2$ $(CPCT)$.
Since $\angle 1$ and $\angle 2$ are alternate interior angles,$EC \parallel BA$.
Now,$EC \parallel BA$ and $BC$ is the transversal,so $\angle ABC + \angle BCE = 180^{\circ}$.
Since $\angle ABC = 90^{\circ}$,$\angle BCE = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
In $\Delta ABC$ and $\Delta ECB$:
$BC = CB$ (Common side)
$AB = EC$ (Proved above)
$\angle ABC = \angle ECB = 90^{\circ}$
Therefore,$\Delta ABC \cong \Delta ECB$ (By $SAS$ congruence criterion).
Thus,$AC = EB$ $(CPCT)$.
Since $BD = DE$,$BD = \frac{1}{2} BE$.
Substituting $BE = AC$,we get $BD = \frac{1}{2} AC$.

(N/A) Given: Lines $l$ and $m$ intersect at point $O$. $P$ is a point on line $n$ passing through $O$ such that $PQ \perp l$ and $PR \perp m$,where $PQ = PR$.
To prove: $n$ is the bisector of $\angle QOR$.
Proof: In $\triangle OQP$ and $\triangle ORP$,we have:
$\angle OQP = \angle ORP = 90^{\circ}$ (Given that $P$ is equidistant from $l$ and $m$)
$OP = OP$ (Common side)
$PQ = PR$ (Given)
By $RHS$ (Right angle-Hypotenuse-Side) criterion of congruence,we have:
$\triangle OQP \cong \triangle ORP$
Therefore,$\angle QOP = \angle ROP$ (by $CPCT$ - Corresponding Parts of Congruent Triangles).
Thus,$n$ is the bisector of $\angle QOR$. Hence proved.

(N/A) Join $MD$ and $MC$.
In $\Delta DMN$ and $\Delta CMN$:
$DN = CN$ [Since $N$ is the mid-point of $DC$]
$\angle DNM = \angle CNM = 90^{\circ}$ [Given]
$MN = MN$ [Common side]
Therefore,$\Delta DMN \cong \Delta CMN$ [By $SAS$ congruence rule]
This implies $DM = CM$ and $\angle NMD = \angle NMC$ ... $(1)$ [$CPCT$]
Now,consider $\Delta AMD$ and $\Delta BMC$:
$AM = BM$ [Since $M$ is the mid-point of $AB$]
$DM = CM$ [From $(1)$]
$\angle AMD = \angle AMN - \angle NMD$
$\angle BMC = \angle BMN - \angle NMC$
Since $\angle AMN = \angle BMN = 90^{\circ}$ and $\angle NMD = \angle NMC$,we have $\angle AMD = \angle BMC$ ... $(2)$
Therefore,$\Delta AMD \cong \Delta BMC$ [By $SAS$ congruence rule]
Thus,$AD = BC$ [$CPCT$].

(N/A) Given: $A$ quadrilateral $ABCD$ such that $\angle 1 = \angle 2$ and $\angle 3 = \angle 4.$
To prove: $AB = AD$ and $CB = CD.$
Proof: In $\triangle ABC$ and $\triangle ADC$,we have:
$\angle 1 = \angle 2$ [Given]
$AC = AC$ [Common side]
$\angle 3 = \angle 4$ [Given]
So,by $ASA$ criterion of congruence,we have:
$\triangle ABC \cong \triangle ADC$
Therefore,$AB = AD$ [by $CPCT$]
And $CB = CD$ [by $CPCT$]
Hence,proved.

(N/A) Given: $A$ right triangle $ABC$ where $AB = AC$ and $CD$ is the bisector of $\angle C$.
To prove: $AC + AD = BC$.
Construction: Draw $DE \perp BC$.
Proof: In right triangle $ABC$,we have $AB = AC$ (Given).
Since $BC$ is the hypotenuse,$\angle A = 90^{\circ}$.
In $\triangle DAC$ and $\triangle DEC$:
$\angle DAC = \angle DEC = 90^{\circ}$ (Given and construction).
$\angle ACD = \angle ECD$ (Since $CD$ is the bisector of $\angle C$).
$CD = CD$ (Common side).
By $AAS$ congruence criterion,$\triangle DAC \cong \triangle DEC$.
Therefore,$AD = DE$ and $AC = CE$ (by $CPCT$).
In $\triangle ABC$,since $AB = AC$,$\angle B = \angle ACB$. Since $\angle A = 90^{\circ}$,$\angle B + \angle ACB = 90^{\circ}$,so $2\angle B = 90^{\circ}$,which means $\angle B = 45^{\circ}$.
In $\triangle DEB$,$\angle DEB = 90^{\circ}$ and $\angle B = 45^{\circ}$,so $\angle EDB = 180^{\circ} - (90^{\circ} + 45^{\circ}) = 45^{\circ}$.
Since $\angle EDB = \angle B = 45^{\circ}$,the sides opposite to them are equal: $DE = BE$.
Since $AD = DE$ and $DE = BE$,we have $AD = BE$.
Now,$BC = CE + BE$.
Substituting $CE = AC$ and $BE = AD$,we get $BC = AC + AD$.
Hence,$AC + AD = BC$ is proved.

(A) Given: $A$ quadrilateral $ABCD$ in which $AB$ is the smallest side and $CD$ is the largest side.
To prove: $\angle B > \angle D$.
Construction: Join $BD$.
Proof: In $\triangle ABD$,since $AB$ is the smallest side of the quadrilateral,we have $AD > AB$. The angle opposite to the longer side is greater,therefore $\angle ABD > \angle ADB$ ... $(1)$.
In $\triangle CBD$,since $CD$ is the largest side of the quadrilateral,we have $CD > BC$. The angle opposite to the longer side is greater,therefore $\angle CBD > \angle BDC$ ... $(2)$.
Adding $(1)$ and $(2)$,we get:
$\angle ABD + \angle CBD > \angle ADB + \angle BDC$
$\Rightarrow \angle ABC > \angle ADC$
$\Rightarrow \angle B > \angle D$.
Hence,it is proved that $\angle B$ is greater than $\angle D$.

(N/A) Given: $A$ triangle $ABC$,which is not an equilateral triangle. Let $BC$ be the longest side.
To prove: $\angle A > \frac{2}{3} \times 90^{\circ} = 60^{\circ}$.
Proof: In $\Delta ABC$,since $BC$ is the longest side,we have:
$BC > AB \Rightarrow \angle A > \angle C$ ..... $(1)$ [Since the angle opposite to the longer side is larger]
$BC > AC \Rightarrow \angle A > \angle B$ ..... $(2)$ [Since the angle opposite to the longer side is larger]
Adding $(1)$ and $(2)$,we get:
$\angle A + \angle A > \angle B + \angle C$
$2\angle A > \angle B + \angle C$
Adding $\angle A$ on both sides:
$2\angle A + \angle A > \angle A + \angle B + \angle C$
$3\angle A > 180^{\circ}$ [Angle sum property of a triangle]
$\angle A > \frac{180^{\circ}}{3}$
$\angle A > 60^{\circ}$
Since $60^{\circ} = \frac{2}{3} \times 90^{\circ}$,we have $\angle A > \frac{2}{3}$ of a right angle.
Hence,proved.

(N/A) Given: $A$ quadrilateral $ABCD$ in which $AB = AD$ and $CB = CD$.
To prove: $AC$ is the perpendicular bisector of $BD$.
Proof: In $\triangle ABC$ and $\triangle ADC$,we have:
$AB = AD$ (Given)
$BC = CD$ (Given)
$AC = AC$ (Common side)
So,by $SSS$ congruence criterion,we have:
$\triangle ABC \cong \triangle ADC$
Therefore,$\angle 1 = \angle 2$ (by $CPCT$)
Now,in $\triangle AOB$ and $\triangle AOD$,we have:
$AB = AD$ (Given)
$\angle 1 = \angle 2$ (Proved above)
$AO = AO$ (Common side)
So,by $SAS$ congruence criterion,we have:
$\triangle AOB \cong \triangle AOD$
Therefore,$BO = DO$ (by $CPCT$)
And $\angle 3 = \angle 4$ (by $CPCT$)
But,$\angle 3 + \angle 4 = 180^{\circ}$ (Linear pair axiom)
Since $\angle 3 = \angle 4$,we have:
$\angle 3 + \angle 3 = 180^{\circ}$
$2\angle 3 = 180^{\circ}$
$\angle 3 = 90^{\circ}$
Since $BO = DO$ and $\angle 3 = 90^{\circ}$,$AC$ is the perpendicular bisector of $BD$.
Hence,proved.

(N/A) Given: $\angle PXR = \angle SXQ$
Subtracting $\angle QXR$ from both sides:
$\angle PXR - \angle QXR = \angle SXQ - \angle QXR$
$\therefore \angle PXQ = \angle SXR \quad \dots(1)$
Now,in $\Delta XPQ$ and $\Delta XSR$:
$XP = XS$ (Given)
$XQ = XR$ (Given)
$\angle PXQ = \angle SXR$ [From $(1)$]
$\therefore \Delta XPQ \cong \Delta XSR$ (By $SAS$ congruence criterion)
$\therefore PQ = SR$ (By $CPCT$)

(N/A) In rectangle $ABCD$,
$AB = DC$ (Opposite sides of a rectangle are equal)
and $\angle B = \angle C = 90^{\circ}$ (Each angle of a rectangle is $90^{\circ}$).
Also,$E$ is the midpoint of $BC$,so $BE = CE$.
Now,consider $\Delta ABE$ and $\Delta DCE$:
$AB = DC$ (Proved above)
$\angle B = \angle C = 90^{\circ}$ (Proved above)
$BE = CE$ (Given that $E$ is the midpoint of $BC$)
By the $SAS$ (Side-Angle-Side) congruence criterion,$\Delta ABE \cong \Delta DCE$.
Since the triangles are congruent,their corresponding parts are equal.
Therefore,$AE = DE$ (by $CPCT$ - Corresponding Parts of Congruent Triangles).

(N/A) Given:
$1$. $PN \perp PQ$ and $QM \perp PQ$.
$2$. $X$ is the midpoint of $PQ$,so $PX = QX$.
$3$. $X$ is the midpoint of $MN$,so $NX = MX$.
To prove: $\triangle PNX \cong \triangle QMX$.
Proof:
In $\triangle PNX$ and $\triangle QMX$:
$1$. $PX = QX$ (Given,$X$ is the midpoint of $PQ$).
$2$. $NX = MX$ (Given,$X$ is the midpoint of $MN$).
$3$. $\angle PNX = \angle QMX$ (Since $PN \parallel QM$ because both are perpendicular to $PQ$,and $MN$ is a transversal,these are alternate interior angles).
Alternatively,using $SAS$ congruence criterion:
$1$. $PX = QX$ (Given).
$2$. $\angle P = \angle Q = 90^\circ$ (Given).
$3$. $NX = MX$ (Given).
Therefore,by $RHS$ congruence criterion (or $SAS$),$\triangle PNX \cong \triangle QMX$.

(N/A) Consider $\triangle APM$ and $\triangle BPN$.
$1$. $\angle MAP = \angle NBP = 90^{\circ}$ (Given that $AM \perp AB$ and $BN \perp AB$).
$2$. $AP = BP$ ($P$ is the midpoint of $AB$,given).
$3$. $\angle APM = \angle BPN$ (Vertically opposite angles).
Therefore,by the $ASA$ (Angle-Side-Angle) congruence criterion,$\triangle APM \cong \triangle BPN$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
- $AM = BN$ (Hence proved).
- $PM = PN$ (Since $PM$ and $PN$ are corresponding sides).
Since $PM = PN$,$P$ is the midpoint of $MN$ (Hence proved).

(N/A) Given: $PQ = ST$,$QU = TR$,$\angle PQT = 90^\circ$ and $\angle STQ = 90^\circ$.
To prove: $PR = SU$.
Proof:
Consider $\triangle PQR$ and $\triangle STU$.
$1$. $PQ = ST$ (Given)
$2$. $\angle PQR = \angle STU = 90^\circ$ (Given)
$3$. $QR = QU + UR$ and $TU = TR + UR$.
Since $QU = TR$,we have $QR = TR + UR = TU$.
Thus,$QR = TU$.
By $SAS$ congruence criterion,$\triangle PQR \cong \triangle STU$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$PR = SU$.

(A) Let $\triangle ABC$ be an equilateral triangle where $AB = BC = CA = a$.
Let $AD$,$BE$,and $CF$ be the medians drawn to sides $BC$,$AC$,and $AB$ respectively.
In an equilateral triangle,the median is also the altitude.
In $\triangle ABD$ and $\triangle BCE$:
$AB = BC = a$ (sides of equilateral triangle).
$\angle B = \angle B$ (common angle).
$BD = BE$ is not the correct approach; let us use the Pythagorean theorem.
In $\triangle ABD$,$\angle ADB = 90^\circ$. By Pythagoras theorem,$AD^2 = AB^2 - BD^2$.
Since $D$ is the midpoint of $BC$,$BD = a/2$.
$AD^2 = a^2 - (a/2)^2 = a^2 - a^2/4 = 3a^2/4$.
So,$AD = (a\sqrt{3})/2$.
Similarly,for median $BE$,in $\triangle BCE$,$BE^2 = BC^2 - CE^2 = a^2 - (a/2)^2 = 3a^2/4$.
So,$BE = (a\sqrt{3})/2$.
Similarly,$CF = (a\sqrt{3})/2$.
Since $AD = BE = CF = (a\sqrt{3})/2$,the medians of an equilateral triangle are equal.

(N/A) Let $PQ$ be a line segment and $L$ be its perpendicular bisector. Let $M$ be a point on $L$ such that $M$ does not lie on $PQ$. Let $L$ intersect $PQ$ at point $O$. Since $L$ is the perpendicular bisector of $PQ$,we have $PO = OQ$ and $\angle MOP = \angle MOQ = 90^{\circ}$. Consider $\triangle MOP$ and $\triangle MOQ$. In these triangles: $1$. $PO = OQ$ (Given). $2$. $\angle MOP = \angle MOQ = 90^{\circ}$ (Given). $3$. $MO = MO$ (Common side). By the $SAS$ (Side-Angle-Side) congruence criterion,$\triangle MOP \cong \triangle MOQ$. Since the triangles are congruent,their corresponding parts are equal $(CPCT)$. Therefore,$MP = MQ$. Thus,point $M$ is equidistant from $P$ and $Q$.

(N/A) Draw the bisector $PM$ of $\angle P$ which meets $QR$ at $M$.
Therefore,$\angle QPM = \angle RPM$ $(1)$.
Now,in $\Delta PMQ$ and $\Delta PMR$:
$PQ = PR$ (Given)
$PM = PM$ (Common side)
$\angle QPM = \angle RPM$ [From $(1)$]
So,by $SAS$ congruence criterion,$\Delta PMQ \cong \Delta PMR$.
Therefore,$\angle R = \angle Q$ (by $CPCT$ - Corresponding Parts of Congruent Triangles).

(N/A) In $\triangle ABC$,$AB = AC$ (Given).
$\therefore \angle ABC = \angle ACB$ ($\because$ Angles opposite to equal sides are equal).
Since $P$ lies on $BC$ and $Q$ lies on $BC$,we have $\angle ABP = \angle ACQ$.
Now,in $\triangle ABP$ and $\triangle ACQ$:
$AB = AC$ (Given)
$BP = CQ$ (Given)
$\angle ABP = \angle ACQ$ (Proved above)
So,by $SAS$ congruence criterion,$\triangle ABP \cong \triangle ACQ$.
$\therefore AP = AQ$ (by $CPCT$).
Now,in $\triangle APQ$,since $AP = AQ$,$\Delta APQ$ is an isosceles triangle.

(A) In $\Delta ABC$,we are given that $AB = AC$.
Since the sides opposite to equal angles are equal,the angles opposite to equal sides are also equal. Therefore,$\angle B = \angle C$.
By the angle sum property of a triangle,the sum of all interior angles is $180^{\circ}$.
So,$\angle A + \angle B + \angle C = 180^{\circ}$.
Given $\angle A = 50^{\circ}$,we have $50^{\circ} + \angle B + \angle B = 180^{\circ}$.
$2 \angle B = 180^{\circ} - 50^{\circ} = 130^{\circ}$.
$\angle B = 130^{\circ} / 2 = 65^{\circ}$.
Since $\angle B = \angle C$,we have $\angle C = 65^{\circ}$.

(A) Given that in $\Delta PQR$,$PQ = PR$.
Since the sides opposite to equal angles are equal,and sides $PQ$ and $PR$ are equal,the angles opposite to these sides must be equal.
Therefore,$\angle R = \angle Q$.
Given $\angle Q = 48^{\circ}$,so $\angle R = 48^{\circ}$.
In any triangle,the sum of all interior angles is $180^{\circ}$.
Thus,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the values: $\angle P + 48^{\circ} + 48^{\circ} = 180^{\circ}$.
$\angle P + 96^{\circ} = 180^{\circ}$.
$\angle P = 180^{\circ} - 96^{\circ} = 84^{\circ}$.
Hence,$\angle P = 84^{\circ}$ and $\angle R = 48^{\circ}$.

(A) Given that in $\Delta ABC$,$AB = AC$. Therefore,the angles opposite to these sides are equal,i.e.,$\angle B = \angle C$.
Since $\angle B = x + 10^{\circ}$,then $\angle C = x + 10^{\circ}$.
The sum of all angles in a triangle is $180^{\circ}$.
So,$\angle A + \angle B + \angle C = 180^{\circ}$.
Substituting the given values: $2x + (x + 10) + (x + 10) = 180$.
$4x + 20 = 180$.
$4x = 160$.
$x = 40$.
Now,calculating the angles:
$\angle A = 2x = 2(40) = 80^{\circ}$.
$\angle B = x + 10 = 40 + 10 = 50^{\circ}$.
$\angle C = x + 10 = 40 + 10 = 50^{\circ}$.

(A) Given that in $\Delta PQR$,$PQ = PR$. Therefore,the angles opposite to these sides are equal,so $\angle Q = \angle R = 4x - 5^{\circ}$.
By the angle sum property of a triangle,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the given values: $(x + 10^{\circ}) + (4x - 5^{\circ}) + (4x - 5^{\circ}) = 180^{\circ}$.
Combining like terms: $9x = 180^{\circ}$,which gives $x = 20^{\circ}$.
Now,calculating the angles:
$\angle P = x + 10^{\circ} = 20^{\circ} + 10^{\circ} = 30^{\circ}$.
$\angle Q = 4x - 5^{\circ} = 4(20^{\circ}) - 5^{\circ} = 80^{\circ} - 5^{\circ} = 75^{\circ}$.
Since $\angle Q = \angle R$,then $\angle R = 75^{\circ}$.
Thus,the angles are $30^{\circ}, 75^{\circ}, 75^{\circ}$.

(N/A) Given: In $\Delta ABC$,$AB = AC$ and $AD \perp BC$.
To prove: $BD = DC$.
Proof: Consider $\Delta ABD$ and $\Delta ACD$.
$1$. $AB = AC$ (Given).
$2$. $\angle ADB = \angle ADC = 90^{\circ}$ (Since $AD$ is an altitude).
$3$. $AD = AD$ (Common side).
By the $RHS$ (Right angle-Hypotenuse-Side) congruence criterion,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = DC$.
This implies that $D$ is the midpoint of $BC$.

(N/A) Given: In $\Delta ABC$,$AB = AC$. $\angle CAD$ is an exterior angle at vertex $A$. $AP$ is the bisector of $\angle CAD$.
To prove: $AP \parallel BC$.
Proof:
$1$. Since $AB = AC$,the angles opposite to equal sides are equal. Therefore,$\angle ABC = \angle ACB$.
$2$. The exterior angle of a triangle is equal to the sum of the two interior opposite angles. Thus,$\angle CAD = \angle ABC + \angle ACB$.
$3$. Since $\angle ABC = \angle ACB$,we can write $\angle CAD = 2 \angle ACB$.
$4$. $AP$ is the bisector of $\angle CAD$,so $\angle CAP = \frac{1}{2} \angle CAD$.
$5$. Substituting the value from step $3$,$\angle CAP = \frac{1}{2} (2 \angle ACB) = \angle ACB$.
$6$. $\angle CAP$ and $\angle ACB$ are alternate interior angles. Since they are equal,the lines $AP$ and $BC$ must be parallel $(AP \parallel BC)$.

(N/A) Given that $BP$ is the bisector of $\angle B$,therefore $\angle XBP = \angle PBC$.
Since $XY \parallel BC$,the alternate interior angles are equal,so $\angle PBC = \angle BXP$.
Thus,$\angle XBP = \angle BXP$.
In $\Delta XBP$,since the base angles are equal,the opposite sides are equal,so $XB = XP$.
Similarly,$CP$ is the bisector of $\angle C$,so $\angle YCP = \angle PCB$.
Since $XY \parallel BC$,the alternate interior angles are equal,so $\angle PCB = \angle CYP$.
Thus,$\angle YCP = \angle CYP$.
In $\Delta YCP$,since the base angles are equal,the opposite sides are equal,so $YC = YP$.
Now,$XY = XP + YP$.
Substituting the values,we get $XY = XB + YC$.
Hence proved.

(N/A) Given: $AB$ and $CD$ bisect each other at $P$. This implies $PA = PB$ and $PC = PD$.
However,the problem states $PA = PD$ and $PB = PC$.
Consider $\triangle APC$ and $\triangle BPD$:
$1$. $PA = PD$ (Given)
$2$. $\angle APC = \angle BPD$ (Vertically opposite angles)
$3$. $PC = PB$ (Given)
By $SAS$ (Side-Angle-Side) congruence criterion,$\triangle APC \cong \triangle BPD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AC = BD$.

(N/A) In $\Delta SPQ$ and $\Delta RQP$:
$PS = QR$ (Given)
$QS = PR$ (Given)
$PQ = QP$ (Common side)
Therefore,by $SSS$ congruence criterion,$\Delta SPQ \cong \Delta RQP$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$(1) \angle PSQ = \angle QRP$ and $(2) \angle SPQ = \angle RQP$.

(N/A) Given: $AB \perp BQ$,$PQ \perp QB$,$AC = PR$ and $BR = QC$.
Since $AB \perp BQ$,$\angle ABQ = 90^{\circ}$. Thus,$\angle ABC = 90^{\circ}$.
Since $PQ \perp QB$,$\angle PQB = 90^{\circ}$. Thus,$\angle PQR = 90^{\circ}$.
Given $BR = QC$.
Adding $RC$ to both sides,we get $BR + RC = QC + RC$.
Therefore,$BC = QR$......$(1)$
In $\Delta ABC$ and $\Delta PQR$:
$AC = PR$ (Given)
$\angle ABC = \angle PQR = 90^{\circ}$ (Proved above)
$BC = QR$ [From $(1)$]
By $RHS$ congruence criterion,$\Delta ABC \cong \Delta PQR$.
Therefore,$\angle BAC = \angle QPR$ (by $CPCT$).

(N/A) Given: In $\Delta PQR$,$X$ is the midpoint of $QR$,so $QX = XR$. $XY \perp PQ$ and $XZ \perp PR$. Also,$XY = XZ$.
Step $1$: Consider $\Delta QXY$ and $\Delta RXZ$.
Step $2$: In these two triangles:
$1$. $\angle XYQ = \angle XZR = 90^{\circ}$ (Given as altitudes).
$2$. $XY = XZ$ (Given).
$3$. $QX = XR$ (Given as $X$ is the midpoint of $QR$).
Step $3$: By $RHS$ congruence criterion,$\Delta QXY \cong \Delta RXZ$.
Step $4$: By $CPCT$,$\angle Q = \angle R$.
Step $5$: In $\Delta PQR$,since $\angle Q = \angle R$,the sides opposite to these angles must be equal,i.e.,$PR = PQ$.
Conclusion: Since two sides of $\Delta PQR$ are equal,$\Delta PQR$ is an isosceles triangle.

(C) parallelogram with equal diagonals is a rectangle.
In a rectangle,all interior angles are equal to $90^{\circ}$.
Therefore,$\angle PQR = 90^{\circ}$.

(N/A) To prove that diagonal $PR$ bisects $\angle QPS$ and $\angle QRS$,we consider the two triangles $\triangle PQR$ and $\triangle PSR$.
$1$. In $\triangle PQR$ and $\triangle PSR$:
- $PQ = PS$ (Given)
- $QR = RS$ (Given)
- $PR = PR$ (Common side)
$2$. By the $SSS$ (Side-Side-Side) congruence criterion,$\triangle PQR \cong \triangle PSR$.
$3$. Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
- $\angle QPR = \angle SPR$. This implies that $PR$ bisects $\angle QPS$.
- $\angle QRP = \angle SRP$. This implies that $PR$ bisects $\angle QRS$.
Thus,it is proved that diagonal $PR$ bisects both $\angle QPS$ and $\angle QRS$.

(N/A) $1$. Consider the two triangles $\triangle PQR$ and $\triangle PSR$.
$2$. In these triangles,we are given that $PQ = PS$ (Side).
$3$. We are also given that $RQ = RS$ (Side).
$4$. The side $PR$ is common to both triangles,so $PR = PR$ (Side).
$5$. By the $SSS$ (Side-Side-Side) congruence criterion,$\triangle PQR \cong \triangle PSR$.
$6$. Since the triangles are congruent,their corresponding parts are equal ($CPCT$ - Corresponding Parts of Congruent Triangles).
$7$. Therefore,$\angle PQR = \angle PSR$.

(N/A) $1$. Consider the right-angled triangles $\Delta PLQ$ and $\Delta QMP$.
$2$. In these triangles,$PL = QM$ (given),$PQ = QP$ (common side),and $\angle PLQ = \angle QMP = 90^{\circ}$ (altitudes).
$3$. By $RHS$ congruence criterion,$\Delta PLQ \cong \Delta QMP$.
$4$. By $CPCT$,$\angle PQL = \angle QPM$,which implies $\angle PQR = \angle QPR$.
$5$. Since the base angles are equal,the sides opposite to them are equal,so $PR = QR$.
$6$. Similarly,by considering $\Delta QMR$ and $\Delta RNP$,we can prove $\Delta QMR \cong \Delta RNP$,which implies $\angle QRM = \angle RPN$,leading to $PQ = QR$.
$7$. Since $PR = QR$ and $PQ = QR$,we have $PQ = QR = PR$.
$8$. Therefore,$\Delta PQR$ is an equilateral triangle.

(N/A) $1$. In $\Delta PQR$,let $PL \perp QR$,$QM \perp PR$,and $RN \perp PQ$. Given $PL = QM = RN$.
$2$. Consider $\Delta QRN$ and $\Delta RQM$. We have $\angle QNR = \angle RMQ = 90^{\circ}$,$QR = RQ$ (common side),and $RN = QM$ (given).
$3$. By $RHS$ congruence,$\Delta QRN \cong \Delta RQM$. Thus,$\angle RQN = \angle QRM$,which implies $\angle RQP = \angle QRP$.
$4$. Similarly,by considering $\Delta PQL$ and $\Delta QPM$,we can show $\angle QPR = \angle PQR$.
$5$. Since $\angle PQR = \angle QRP$ and $\angle QPR = \angle PQR$,it follows that $\angle PQR = \angle QRP = \angle QPR$.
$6$. Since all angles are equal,$\Delta PQR$ is an equilateral triangle.

(N/A) In $\Delta XYZ$,$XY > XZ$ (Given).
$\therefore \angle XZY > \angle XYZ$ (Since the angle opposite to the larger side of a triangle is greater).
In $\Delta XPZ$,$\angle XPY$ is an exterior angle,and $\angle XZP$ is its interior opposite angle.
Therefore,$\angle XPY > \angle XZP$ (Exterior angle property).
Since $\angle XZY$ is the same as $\angle XZP$,we have $\angle XPY > \angle XZP > \angle XYZ$.
Thus,$\angle XPY > \angle XYZ$ (or $\angle XPY > \angle XYP$).
In $\Delta XYP$,since the angle opposite to $XY$ (which is $\angle XPY$) is greater than the angle opposite to $XP$ (which is $\angle XYP$),it follows that $XY > XP$.

(A) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
Therefore,$\angle C = 180^{\circ} - (50^{\circ} + 60^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
We know that the side opposite to the smaller angle is smaller and the side opposite to the larger angle is larger.
The angles in ascending order are $\angle A < \angle B < \angle C$ $(50^{\circ} < 60^{\circ} < 70^{\circ})$.
Therefore,the sides in ascending order are $BC < AC < AB$.

(QR, PR, PQ) In $\Delta PQR$,the sum of all interior angles is $180^{\circ}$.
Therefore,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the given values: $40^{\circ} + \angle Q + 80^{\circ} = 180^{\circ}$.
$120^{\circ} + \angle Q = 180^{\circ}$,which gives $\angle Q = 60^{\circ}$.
We know that the side opposite to the smallest angle is the shortest,and the side opposite to the largest angle is the longest.
The angles in ascending order are $\angle P (40^{\circ}) < \angle Q (60^{\circ}) < \angle R (80^{\circ})$.
Therefore,the sides opposite to these angles in ascending order are $QR < PR < PQ$.

(A) In $\Delta XYZ$,the sum of angles is $180^{\circ}$.
Therefore,$\angle Z = 180^{\circ} - (80^{\circ} + 30^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
The angles in ascending order are $\angle Y (30^{\circ}) < \angle Z (70^{\circ}) < \angle X (80^{\circ})$.
Since the side opposite to the smaller angle is smaller,the sides in ascending order are $XZ < XY < YZ$.

(N/A) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
$\angle A + 70^{\circ} + 20^{\circ} = 180^{\circ}$.
$\angle A + 90^{\circ} = 180^{\circ}$,so $\angle A = 90^{\circ}$.
The angles are $\angle C = 20^{\circ}$,$\angle B = 70^{\circ}$,and $\angle A = 90^{\circ}$.
Since the side opposite to the smaller angle is smaller,the order of sides is $AB < AC < BC$.

(A) In $\Delta PQR$,the sum of all angles is $180^{\circ}$.
$\angle P + \angle Q + \angle R = 180^{\circ}$
$93^{\circ} + \angle Q + 55^{\circ} = 180^{\circ}$
$148^{\circ} + \angle Q = 180^{\circ}$
$\angle Q = 180^{\circ} - 148^{\circ} = 32^{\circ}$.
Since the side opposite to the smallest angle is the shortest and the side opposite to the largest angle is the longest,we compare the angles: $\angle Q (32^{\circ}) < \angle R (55^{\circ}) < \angle P (93^{\circ})$.
Therefore,the sides in ascending order are: $PR < PQ < QR$.

(A) In a triangle,the angle opposite to the longer side is larger.
Given the side lengths are in the order $AB > BC > CA$.
The angle opposite to side $AB$ is $\angle C$.
The angle opposite to side $BC$ is $\angle A$.
The angle opposite to side $CA$ is $\angle B$.
Since $AB > BC > CA$,it follows that $\angle C > \angle A > \angle B$.
Therefore,the angles in ascending order are $\angle B < \angle A < \angle C$.

(N/A) Let $\angle ABC = \beta$ and $\angle ACB = \gamma$.
Since $\angle ABD$ is an exterior angle at vertex $B$,$\angle ABD = 180^\circ - \beta$.
Since $\angle ACE$ is an exterior angle at vertex $C$,$\angle ACE = 180^\circ - \gamma$.
Given that $\angle ABD > \angle ACE$,we substitute the expressions:
$180^\circ - \beta > 180^\circ - \gamma$.
Subtracting $180^\circ$ from both sides,we get $-\beta > -\gamma$.
Multiplying by $-1$ reverses the inequality sign: $\beta < \gamma$.
Since $\angle ACB (\gamma) > \angle ABC (\beta)$,the side opposite to the larger angle is longer.
Therefore,the side opposite to $\angle ACB$ is $AB$ and the side opposite to $\angle ABC$ is $AC$.
Thus,$AB > AC$ is incorrect based on the standard exterior angle theorem logic; if $\angle ABD > \angle ACE$,then $\angle ACB > \angle ABC$,which implies $AB > AC$ is false,rather $AC > AB$ is true.

(N/A) Let the sides of the scalene triangle be $a$,$b$,and $c$ such that $a > b > c$.
Let the angles opposite to these sides be $A$,$B$,and $C$ respectively.
According to the property of triangles,the angle opposite to the longest side is the largest angle. Thus,$A > B > C$.
We know that the sum of all angles in a triangle is $180^{\circ}$,so $A + B + C = 180^{\circ}$.
Since $A > B$ and $A > C$,we can write $A + A + A > A + B + C$.
Therefore,$3A > 180^{\circ}$.
Dividing both sides by $3$,we get $A > 60^{\circ}$.
Thus,the angle opposite to the longest side in a scalene triangle is always greater than $60^{\circ}$.

(N/A) In $\Delta ABD$,by the triangle inequality theorem,the sum of any two sides is greater than the third side: $AB + BD > AD$ (Equation $1$).
In $\Delta ACD$,by the triangle inequality theorem: $AC + CD > AD$ (Equation $2$).
Adding Equation $1$ and Equation $2$,we get: $(AB + BD) + (AC + CD) > AD + AD$.
Rearranging the terms: $AB + AC + (BD + CD) > 2 AD$.
Since $D$ is a point on $BC$,$BD + CD = BC$.
Therefore,$AB + AC + BC > 2 AD$.
This shows that the perimeter of $\Delta ABC$ is greater than $2 AD$.