Textbook - Triangles Questions in English

(N/A) $(i)$ In $\Delta AOD$ and $\Delta BOC$:
$OA = OB$ (Given)
$OD = OC$ (Given)
Also,$\angle AOD = \angle BOC$ (Vertically opposite angles).
Therefore,by the $SAS$ congruence rule,$\Delta AOD \cong \Delta BOC$.
$(ii)$ Since $\Delta AOD \cong \Delta BOC$,their corresponding parts are equal $(CPCT)$.
Thus,$\angle OAD = \angle OBC$.
These are alternate interior angles formed by the transversal $AB$ intersecting lines $AD$ and $BC$.
Since the alternate interior angles are equal,$AD \parallel BC$.

(N/A) Line $l$ is the perpendicular bisector of $AB$ and passes through $C$,which is the mid-point of $AB$ (see figure).
We need to show that $PA = PB$.
Consider $\Delta PCA$ and $\Delta PCB$.
In these triangles:
$AC = BC$ ($C$ is the mid-point of $AB$)
$\angle PCA = \angle PCB = 90^\circ$ (Given,as $l$ is the perpendicular bisector)
$PC = PC$ (Common side)
Therefore,by the $SAS$ congruence rule,$\Delta PCA \cong \Delta PCB$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$PA = PB$,which means $P$ is equidistant from $A$ and $B$.

$(i)$ Consider $\Delta AOB$ and $\Delta DOC$.
$\angle OAB = \angle ODC$ (Alternate interior angles as $AB \parallel CD$ and $AD$ is the transversal)
$\angle AOB = \angle DOC$ (Vertically opposite angles)
$OA = OD$ (Given,as $O$ is the mid-point of $AD$)
Therefore,by $ASA$ congruence rule,$\Delta AOB \cong \Delta DOC$.
$(ii)$ Since $\Delta AOB \cong \Delta DOC$,then by $CPCT$ (Corresponding Parts of Congruent Triangles),
$OB = OC$
Thus,$O$ is the mid-point of $BC$.

(N/A) In quadrilateral $ACBD$,we are given:
$AC = AD$
$AB$ bisects $\angle A$,which implies $\angle CAB = \angle DAB$.
Now,consider $\Delta ABC$ and $\Delta ABD$:
$1$. $AC = AD$ (Given)
$2$. $\angle CAB = \angle DAB$ (Since $AB$ bisects $\angle A$)
$3$. $AB = AB$ (Common side)
By the $SAS$ (Side-Angle-Side) congruence criterion,$\Delta ABC \cong \Delta ABD$.
Since the triangles are congruent,their corresponding parts are equal by $CPCT$ (Corresponding Parts of Congruent Triangles).
Therefore,$BC = BD$.

(N/A) $(i)$ In quadrilateral $ABCD$,we have $AD = BC$ and $\angle DAB = \angle CBA$.
Consider $\Delta ABD$ and $\Delta BAC$:
$AD = BC$ (Given)
$AB = BA$ (Common side)
$\angle DAB = \angle CBA$ (Given)
Therefore,by $SAS$ (Side-Angle-Side) congruence criterion,we have $\Delta ABD \cong \Delta BAC$.
$(ii)$ Since $\Delta ABD \cong \Delta BAC$,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = AC$.
$(iii)$ Since $\Delta ABD \cong \Delta BAC$,their corresponding parts are equal $(CPCT)$.
Therefore,$\angle ABD = \angle BAC$.

(N/A) Given: $AD \perp AB$,$BC \perp AB$ and $AD = BC$.
To prove: $CD$ bisects $AB$,i.e.,$OA = OB$.
Proof:
In $\Delta OBC$ and $\Delta OAD$:
$1$. $\angle OBC = \angle OAD = 90^\circ$ (Given)
$2$. $\angle BOC = \angle AOD$ (Vertically opposite angles)
$3$. $BC = AD$ (Given)
By $AAS$ congruence rule,$\Delta OBC \cong \Delta OAD$.
Therefore,$OB = OA$ (by $c.p.c.t.$).
Since $OB = OA$,$O$ is the midpoint of $AB$.
Hence,$CD$ bisects $AB$.

(N/A) Given: $l \parallel m$ and $p \parallel q$.
To prove: $\Delta ABC \cong \Delta CDA$.
Proof:
Since $l \parallel m$ and $AC$ is a transversal,
$\therefore \angle BAC = \angle DCA$ [Alternate interior angles]
Also,since $p \parallel q$ and $AC$ is a transversal,
$\therefore \angle BCA = \angle DAC$ [Alternate interior angles]
Now,in $\Delta ABC$ and $\Delta CDA$:
$\angle BAC = \angle DCA$ [Proved above]
$\angle BCA = \angle DAC$ [Proved above]
$AC = CA$ [Common side]
Therefore,by the $ASA$ (Angle-Side-Angle) congruence criterion,$\Delta ABC \cong \Delta CDA$.

(N/A) Given: Line $l$ is the bisector of $\angle A$. $BP \perp AQ$ and $BQ \perp AP$.
$(i)$ In $\Delta APB$ and $\Delta AQB$:
$\angle APB = \angle AQB = 90^{\circ}$ (Given)
$\angle PAB = \angle QAB$ ($l$ is the bisector of $\angle A$)
$AB = AB$ (Common side)
Therefore,by $AAS$ congruence criterion,$\Delta APB \cong \Delta AQB$.
$(ii)$ Since $\Delta APB \cong \Delta AQB$,their corresponding parts are equal by $CPCT$.
Therefore,$BP = BQ$.
This shows that point $B$ is equidistant from the arms of $\angle A$.

(N/A) We have $\angle BAD = \angle EAC$.
Adding $\angle DAC$ on both sides,we have:
$\angle BAD + \angle DAC = \angle EAC + \angle DAC$
$\Rightarrow \angle BAC = \angle DAE$
Now,in $\triangle ABC$ and $\triangle ADE$,we have:
$\angle BAC = \angle DAE$ [Proved above]
$AB = AD$ [Given]
$AC = AE$ [Given]
$\therefore \triangle ABC \cong \triangle ADE$ [Using $SAS$ congruence criterion]
Since $\triangle ABC \cong \triangle ADE$,therefore,their corresponding parts are equal $(CPCT)$.
$\Rightarrow BC = DE$.

(N/A) Given: $P$ is the mid-point of $AB$.
$\therefore AP = BP$
$\angle EPA = \angle DPB$ [Given]
Adding $\angle EPD$ to both sides,we get:
$\angle EPA + \angle EPD = \angle DPB + \angle EPD$
$\Rightarrow \angle APD = \angle BPE$
$(i)$ In $\Delta DAP$ and $\Delta EBP$:
$AP = BP$ [Proved above]
$\angle PAD = \angle PBE$ [Given as $\angle BAD = \angle ABE$]
$\angle APD = \angle BPE$ [Proved above]
By $ASA$ congruence criterion,we have:
$\Delta DAP \cong \Delta EBP$
$(ii)$ Since $\Delta DAP \cong \Delta EBP$,their corresponding parts are equal $(CPCT)$:
$\Rightarrow AD = BE$

(N/A) Given: In $\Delta ABC$,$\angle C = 90^\circ$ and $M$ is the mid-point of $AB$ $(AM = BM)$. $CM$ is produced to $D$ such that $DM = CM$.
$(i)$ In $\Delta AMC$ and $\Delta BMD$:
$AM = BM$ (Given,$M$ is the mid-point of $AB$)
$\angle AMC = \angle BMD$ (Vertically opposite angles)
$CM = DM$ (Given)
Therefore,by $SAS$ congruence criterion,$\Delta AMC \cong \Delta BMD$.
$(ii)$ Since $\Delta AMC \cong \Delta BMD$,their corresponding parts are equal $(c.p.c.t.)$.
So,$\angle MAC = \angle MBD$. These are alternate interior angles,which implies $AC \parallel DB$.
Since $AC \parallel DB$ and $BC$ is a transversal,the sum of consecutive interior angles is $180^\circ$.
$\angle ACB + \angle DBC = 180^\circ$
$90^\circ + \angle DBC = 180^\circ$
$\angle DBC = 180^\circ - 90^\circ = 90^\circ$.
$(iii)$ In $\Delta DBC$ and $\Delta ACB$:
$DB = AC$ ($c.p.c.t.$ from $\Delta AMC \cong \Delta BMD$)
$\angle DBC = \angle ACB = 90^\circ$ (Proved above)
$BC = CB$ (Common side)
Therefore,by $SAS$ congruence criterion,$\Delta DBC \cong \Delta ACB$.
$(iv)$ Since $\Delta DBC \cong \Delta ACB$,their corresponding parts are equal $(c.p.c.t.)$.
So,$DC = AB$.
Since $DM = CM$,$CM = \frac{1}{2} DC$.
Substituting $DC = AB$,we get $CM = \frac{1}{2} AB$.

(N/A) In $\Delta ABD$ and $\Delta ACD$:
$\angle BAD = \angle CAD$ (Given,as $AD$ is the bisector of $\angle A$)
$AD = AD$ (Common side)
$\angle ADB = \angle ADC = 90^\circ$ (Given,as $AD \perp BC$)
Therefore,by the $ASA$ (Angle-Side-Angle) congruence rule,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AB = AC$.
Since two sides of $\Delta ABC$ are equal,$\Delta ABC$ is an isosceles triangle.

(N/A) In $\Delta ABF$ and $\Delta ACE$:
$AB = AC$ (Given)
$\angle A = \angle A$ (Common angle)
$AF = AE$ (Since $F$ and $E$ are mid-points of equal sides $AC$ and $AB$ respectively,$AF = \frac{1}{2} AC$ and $AE = \frac{1}{2} AB$. Since $AB = AC$,it follows that $AF = AE$.)
Therefore,$\Delta ABF \cong \Delta ACE$ by the $SAS$ congruence rule.
Since the triangles are congruent,their corresponding parts are equal.
Thus,$BF = CE$ (by $CPCT$).

(N/A) In $\Delta ABD$ and $\Delta ACE$:
$AB = AC$ (Given) ........... $(1)$
$\angle B = \angle C$ (Angles opposite to equal sides) ........... $(2)$
Also,$BE = CD$
Subtracting $DE$ from both sides:
$BE - DE = CD - DE$
$BD = CE$ ........... $(3)$
Therefore,$\Delta ABD \cong \Delta ACE$ (Using $(1)$,$(2)$,$(3)$ and $SAS$ congruence rule).
This gives $AD = AE$ $(CPCT)$.

(N/A) $(i)$ In $\Delta ABC$,we have
$AB = AC$ [Given]
$\therefore \angle C = \angle B$ [Angles opposite to equal sides are equal]
$\Rightarrow \frac{1}{2} \angle C = \frac{1}{2} \angle B$
Or $\angle OCB = \angle OBC$
$\Rightarrow OB = OC$ [Sides opposite to equal angles are equal]
$(ii)$ In $\Delta ABO$ and $\Delta ACO$,we have
$AB = AC$ [Given]
$OB = OC$ [Proved above]
$\angle OBA = \angle OCA$ [Since $\frac{1}{2} \angle B = \frac{1}{2} \angle C$]
$\therefore$ Using $SAS$ congruence criteria,
$\Delta ABO \cong \Delta ACO$
$\Rightarrow \angle OAB = \angle OAC$ [c.p.c.t.]
$\Rightarrow AO$ bisects $\angle A$.

(N/A) Given: $AD$ is the perpendicular bisector of $BC$,which implies $BD = CD$ and $\angle ADB = \angle ADC = 90^o$.
In $\Delta ABD$ and $\Delta ACD$:
$AD = AD$ (Common side)
$\angle ADB = \angle ADC = 90^o$ (Given)
$BD = CD$ (Since $AD$ is the bisector of $BC$)
Therefore,$\Delta ABD \cong \Delta ACD$ by the $SAS$ congruence criterion.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AB = AC$.
Since two sides of $\Delta ABC$ are equal,it is an isosceles triangle.

(N/A) Given: $\Delta ABC$ is an isosceles triangle with $AB = AC$.
Altitudes $BE \perp AC$ and $CF \perp AB$.
To prove: $BE = CF$.
Proof:
In $\Delta ABE$ and $\Delta ACF$:
$1$. $\angle A = \angle A$ (Common angle)
$2$. $AB = AC$ (Given)
$3$. $\angle AEB = \angle AFC = 90^\circ$ (Since $BE$ and $CF$ are altitudes)
Therefore,by $AAS$ congruence criterion,$\Delta ABE \cong \Delta ACF$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$BE = CF$.

(N/A) $(i)$ In $\Delta ABE$ and $\Delta ACF$,we have:
$\angle AEB = \angle AFC = 90^\circ$ (Since $BE \perp AC$ and $CF \perp AB$)
$\angle A = \angle A$ (Common angle)
$BE = CF$ (Given)
Therefore,by the $AAS$ congruence criterion,$\Delta ABE \cong \Delta ACF$.
$(ii)$ Since $\Delta ABE \cong \Delta ACF$,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$.
Thus,$ABC$ is an isosceles triangle.

(N/A) In $\Delta ABC$,we have
$AB = AC$ (Given,since $\Delta ABC$ is an isosceles triangle)
Since angles opposite to equal sides are equal,we have:
$\angle ABC = \angle ACB$ .......... $(1)$
In $\Delta BDC$,we have
$BD = CD$ (Given,since $\Delta BDC$ is an isosceles triangle)
Since angles opposite to equal sides are equal,we have:
$\angle CBD = \angle BCD$ .......... $(2)$
Adding equations $(1)$ and $(2)$,we get:
$\angle ABC + \angle CBD = \angle ACB + \angle BCD$
$\Rightarrow \angle ABD = \angle ACD$
Hence proved.

(N/A) In $\Delta ABC$,since $AB = AC$,we have $\angle ABC = \angle ACB$ (angles opposite to equal sides). Let $\angle ABC = \angle ACB = x$. Then $\angle BAC = 180^\circ - 2x$.
In $\Delta ACD$,since $AD = AB$ and $AB = AC$,we have $AD = AC$. Thus,$\angle ADC = \angle ACD$ (angles opposite to equal sides). Let $\angle ADC = \angle ACD = y$.
Since $BA$ is produced to $D$,$BD$ is a straight line. Therefore,$\angle BAC + \angle CAD = 180^\circ$.
Since $\angle CAD = 180^\circ - 2y$ (from $\Delta ACD$),we have $(180^\circ - 2x) + (180^\circ - 2y) = 180^\circ$.
$360^\circ - 2(x + y) = 180^\circ$.
$2(x + y) = 180^\circ$.
$x + y = 90^\circ$.
Since $\angle BCD = \angle ACB + \angle ACD = x + y$,we conclude $\angle BCD = 90^\circ$.

(N/A) In $\triangle ABC$,we have:
$AB = AC$ [Given]
Since the sides are equal,their opposite angles are also equal.
$\Rightarrow \angle ACB = \angle ABC$
We know that the sum of all angles in a triangle is $180^o$.
$\angle A + \angle B + \angle C = 180^o$
$90^o + \angle B + \angle C = 180^o$ [Since $\angle A = 90^o$]
$\angle B + \angle C = 180^o - 90^o = 90^o$
Since $\angle B = \angle C$,we can write:
$2 \angle B = 90^o$
$\angle B = \frac{90^o}{2} = 45^o$
Therefore,$\angle B = 45^o$ and $\angle C = 45^o$.

(N/A) In $\Delta ABC$,we have:
$AB = BC = CA$ $[\because \Delta ABC$ is an equilateral triangle$]$
Since $AB = BC$,the angles opposite to these sides are equal,so $\angle A = \angle C$ ......... $(1)$
Similarly,since $AC = BC$,the angles opposite to these sides are equal,so $\angle A = \angle B$ ......... $(2)$
From $(1)$ and $(2)$,we have:
$\angle A = \angle B = \angle C$
Let $\angle A = \angle B = \angle C = x$
Since the sum of the angles in a triangle is $180^o$:
$\angle A + \angle B + \angle C = 180^o$
Substituting $x$ for each angle:
$x + x + x = 180^o$
$3x = 180^o$
$x = \frac{180^o}{3} = 60^o$
Therefore,$\angle A = \angle B = \angle C = 60^o$.
Thus,each angle of an equilateral triangle is $60^o$.

(N/A) Given: $PA = PB$ and $QA = QB$. We need to show that $PQ \perp AB$ and $PQ$ bisects $AB$. Let $PQ$ intersect $AB$ at $C$.
Consider $\Delta PAQ$ and $\Delta PBQ$:
$AP = BP$ (Given)
$AQ = BQ$ (Given)
$PQ = PQ$ (Common side)
Therefore,$\Delta PAQ \cong \Delta PBQ$ by $SSS$ congruence rule.
This implies $\angle APQ = \angle BPQ$ by $CPCT$.
Now,consider $\Delta PAC$ and $\Delta PBC$:
$AP = BP$ (Given)
$\angle APC = \angle BPC$ (Since $\angle APQ = \angle BPQ$)
$PC = PC$ (Common side)
Therefore,$\Delta PAC \cong \Delta PBC$ by $SAS$ congruence rule.
This implies $AC = BC$ and $\angle ACP = \angle BCP$ by $CPCT$.
Since $\angle ACP + \angle BCP = 180^o$ (Linear pair),
$2 \angle ACP = 180^o \implies \angle ACP = 90^o$.
Since $AC = BC$ and $\angle ACP = 90^o$,$PQ$ is the perpendicular bisector of $AB$.

(N/A) Given: Lines $l$ and $m$ intersect at point $A$. $P$ is a point such that $PB \perp l$ and $PC \perp m$,with $PB = PC$.
To prove: Line $AP$ bisects the angle between lines $l$ and $m$,i.e.,$\angle PAB = \angle PAC$.
Proof: Consider $\Delta PAB$ and $\Delta PAC$.
In these two triangles:
$1$. $PB = PC$ (Given,as $P$ is equidistant from the lines)
$2$. $\angle PBA = \angle PCA = 90^o$ (Given,as $PB \perp l$ and $PC \perp m$)
$3$. $PA = PA$ (Common side)
Therefore,by the $RHS$ congruence rule,$\Delta PAB \cong \Delta PAC$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$\angle PAB = \angle PAC$.
This shows that the line $AP$ bisects the angle between lines $l$ and $m$.

(N/A) $(i)$ In $\Delta ABD$ and $\Delta ACD$,we have:
$AB = AC$ [Given]
$AD = AD$ [Common]
$BD = CD$ [Given]
$\therefore \Delta ABD \cong \Delta ACD$ [$SSS$ congruence criterion]
$(ii)$ In $\Delta ABP$ and $\Delta ACP$,we have:
$AB = AC$ [Given]
$\angle BAP = \angle CAP$ [Since $\Delta ABD \cong \Delta ACD$,their corresponding parts are equal]
$AP = AP$ [Common]
$\therefore \Delta ABP \cong \Delta ACP$ [$SAS$ congruence criterion]
$(iii)$ Since $\Delta ABP \cong \Delta ACP$,their corresponding parts are congruent.
$\Rightarrow \angle BAP = \angle CAP$,so $AP$ bisects $\angle A$.
Also,in $\Delta BDP$ and $\Delta CDP$:
$BD = CD$ [Given]
$DP = DP$ [Common]
$BP = CP$ [Since $\Delta ABP \cong \Delta ACP$,$BP = CP$ by $CPCT$]
$\therefore \Delta BDP \cong \Delta CDP$ [$SSS$ congruence criterion]
$\Rightarrow \angle BDP = \angle CDP$,so $AP$ bisects $\angle D$.
$(iv)$ Since $\Delta ABP \cong \Delta ACP$,we have $\angle APB = \angle APC$ [by $CPCT$].
Since $\angle APB + \angle APC = 180^\circ$ [Linear pair],
$\angle APB = \angle APC = 90^\circ$.
Also $BP = CP$ [by $CPCT$].
Thus,$AP$ is the perpendicular bisector of $BC$.

(N/A) $(i)$ In $\Delta ABD$ and $\Delta ACD$,we have:
$AB = AC$ [Given]
$\angle ADB = \angle ADC = 90^\circ$ [Since $AD$ is an altitude]
$AD = AD$ [Common side]
Therefore,by $RHS$ congruence rule,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = CD$,which means $AD$ bisects $BC$.
$(ii)$ Since $\Delta ABD \cong \Delta ACD$,their corresponding angles are equal $(CPCT)$.
Therefore,$\angle BAD = \angle CAD$,which means $AD$ bisects $\angle A$.

(N/A) In $\Delta ABC$,$AM$ is a median [Given].
$\therefore BM = \frac{1}{2} BC$ ........... $(1)$
In $\Delta PQR$,$PN$ is a median.
$\therefore QN = \frac{1}{2} QR$ ........... $(2)$
$\because BC = QR$ [Given]
$\therefore \frac{1}{2} BC = \frac{1}{2} QR$
$\Rightarrow BM = QN$ [From $(1)$ and $(2)$]
$(i)$ In $\Delta ABM$ and $\Delta PQN$,we have:
$AB = PQ$ [Given]
$AM = PN$ [Given]
$BM = QN$ [Proved above]
$\therefore \Delta ABM \cong \Delta PQN$ [$SSS$ congruence criterion]
$(ii)$ $\because \Delta ABM \cong \Delta PQN$
$\therefore$ Their corresponding parts are congruent $(CPCT)$.
$\Rightarrow \angle B = \angle Q$
Now,in $\Delta ABC$ and $\Delta PQR$,we have:
$AB = PQ$ [Given]
$\angle B = \angle Q$ [Proved above]
$BC = QR$ [Given]
$\therefore \Delta ABC \cong \Delta PQR$ [$SAS$ congruence criterion]

(N/A) Given: $BE \perp AC$ and $CF \perp AB$ such that $BE = CF$.
To prove: $\Delta ABC$ is an isosceles triangle,i.e.,$AB = AC$.
Proof:
In right-angled triangles $\Delta BEC$ and $\Delta CFB$:
$1$. $\angle BEC = \angle CFB = 90^\circ$ (Altitudes are perpendicular to the sides).
$2$. $BC = CB$ (Common hypotenuse).
$3$. $BE = CF$ (Given).
Therefore,by $RHS$ congruence rule,$\Delta BEC \cong \Delta CFB$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
$\angle BCE = \angle CBF$
This implies $\angle BCA = \angle CBA$.
In $\Delta ABC$,since the angles opposite to equal sides are equal,the sides opposite to equal angles are also equal.
Therefore,$AB = AC$.
Hence,$\Delta ABC$ is an isosceles triangle.

(N/A) Given: $ABC$ is an isosceles triangle with $AB = AC$. $AP \perp BC$.
To prove: $\angle B = \angle C$.
Proof:
In $\Delta ABP$ and $\Delta ACP$:
$1$. $\angle APB = \angle APC = 90^\circ$ (Since $AP \perp BC$)
$2$. $AB = AC$ (Given)
$3$. $AP = AP$ (Common side)
Therefore,by $RHS$ congruence criterion,$\Delta ABP \cong \Delta ACP$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$\angle B = \angle C$.

(N/A) In $\Delta DAC$,
$AD = AC$ (Given)
Therefore,$\angle ADC = \angle ACD$ (Angles opposite to equal sides are equal).
Now,$\angle ADC$ is an exterior angle for $\Delta ABD$.
By the exterior angle theorem,$\angle ADC > \angle ABD$.
Since $\angle ADC = \angle ACD$,we have $\angle ACD > \angle ABD$.
This implies $\angle ACB > \angle ABC$.
In $\Delta ABC$,the side opposite to the larger angle is longer. Therefore,$AB > AC$.
Since $AD = AC$,we conclude that $AB > AD$.

(N/A) Let us consider $\Delta ABC$ such that $\angle B = 90^\circ$.
Since the sum of angles in a triangle is $180^\circ$,we have $\angle A + \angle B + \angle C = 180^\circ$.
Substituting $\angle B = 90^\circ$,we get $\angle A + 90^\circ + \angle C = 180^\circ$,which implies $\angle A + \angle C = 90^\circ$.
Since $\angle A + \angle C = 90^\circ$ and $\angle B = 90^\circ$,it follows that $\angle B = \angle A + \angle C$.
This implies that $\angle B > \angle A$ and $\angle B > \angle C$.
In any triangle,the side opposite to the larger angle is longer. Therefore,the side opposite to $\angle B$ (which is $AC$) is longer than the side opposite to $\angle A$ (which is $BC$) and the side opposite to $\angle C$ (which is $AB$).
Thus,$AC > BC$ and $AC > AB$.
Since $AC$ is the hypotenuse,it is the longest side of the triangle.

(N/A) We know that $\angle ABC + \angle PBC = 180^\circ$ [Linear pair axiom].
Similarly,$\angle ACB + \angle QCB = 180^\circ$ [Linear pair axiom].
Therefore,$\angle ABC + \angle PBC = \angle ACB + \angle QCB$.
Given that $\angle PBC < \angle QCB$.
Since the sum of the angles is constant $(180^\circ)$,if one angle in the sum is smaller,the other must be larger to maintain the equality.
Thus,$\angle ABC > \angle ACB$.
In any triangle,the side opposite to the larger angle is longer.
Therefore,the side opposite to $\angle ABC$ (which is $AC$) is greater than the side opposite to $\angle ACB$ (which is $AB$).
Hence,$AC > AB$.

(N/A) Given: $\angle B < \angle A$ and $\angle C < \angle D$.
In $\triangle AOB$,since $\angle B < \angle A$,the side opposite to the greater angle is longer. Therefore,$OA < OB$ (side opposite to $\angle B$ is $OA$ and side opposite to $\angle A$ is $OB$) ...... $(1)$
In $\triangle COD$,since $\angle C < \angle D$,the side opposite to the greater angle is longer. Therefore,$OD < OC$ (side opposite to $\angle C$ is $OD$ and side opposite to $\angle D$ is $OC$) ...... $(2)$
Adding equations $(1)$ and $(2)$,we get:
$OA + OD < OB + OC$
Since $OA + OD = AD$ and $OB + OC = BC$,we have:
$AD < BC$.

(N/A) Let us join $AC$.
In $\Delta ABC$,since $AB$ is the smallest side of the quadrilateral $ABCD$,we have $AB < BC$ and $AB < AC$.
Specifically,$BC > AB$.
Since the angle opposite to the longer side is larger,we have $\angle BAC > \angle BCA$ ........... $(1)$
In $\Delta ACD$,since $CD$ is the longest side of the quadrilateral $ABCD$,we have $CD > AD$ and $CD > AC$.
Specifically,$CD > AD$.
Since the angle opposite to the longer side is larger,we have $\angle CAD > \angle ACD$ ........... $(2)$
Adding $(1)$ and $(2)$,we get:
$(\angle BAC + \angle CAD) > (\angle BCA + \angle ACD)$
$\Rightarrow \angle A > \angle C$
Similarly,by joining $BD$ and using the same logic for $\Delta ABD$ and $\Delta BCD$,we can show that $\angle B > \angle D$.

(N/A) In $\Delta PQR$,$PS$ bisects $\angle QPR$ [Given].
Therefore,$\angle QPS = \angle RPS$.
Since $PR > PQ$ [Given],
Therefore,the angle opposite to $PR$ is greater than the angle opposite to $PQ$.
Thus,$\angle PQS > \angle PRS$.
Adding $\angle QPS$ to the left side and $\angle RPS$ to the right side (since $\angle QPS = \angle RPS$):
$\angle PQS + \angle QPS > \angle PRS + \angle RPS$ ... $(1)$.
We know that the exterior angle of a triangle is equal to the sum of the two interior opposite angles.
For $\Delta PQS$,the exterior angle $\angle PSR = \angle PQS + \angle QPS$.
For $\Delta PRS$,the exterior angle $\angle PSQ = \angle PRS + \angle RPS$.
Substituting these into $(1)$,we get $\angle PSR > \angle PSQ$.

(N/A) Let us consider a point $P$ not on line $l$. Let $PM$ be the perpendicular from $P$ to line $l$,and let $N$ be any other point on line $l$ such that $N \neq M$.
In $\Delta PMN$,since $\angle M = 90^o$,the sum of angles is $\angle M + \angle N + \angle P = 180^o$.
This implies $\angle N + \angle P = 90^o$,so $\angle N < 90^o$.
Since $\angle N < \angle M$,the side opposite to $\angle N$ must be smaller than the side opposite to $\angle M$.
Therefore,$PM < PN$.
Since $N$ is an arbitrary point on line $l$ (other than $M$),this shows that $PM$ is shorter than any other line segment drawn from $P$ to line $l$.
Thus,the perpendicular segment is the shortest line segment drawn from a point to a line.

(N/A) Let us consider a $\Delta ABC$.
Draw line $l$ as the perpendicular bisector of side $AB$.
Draw line $m$ as the perpendicular bisector of side $BC$.
Let the two perpendicular bisectors $l$ and $m$ intersect at point $O$.
Since any point on the perpendicular bisector of a segment is equidistant from its endpoints,point $O$ lies on the perpendicular bisector of $AB$ (so $OA = OB$) and on the perpendicular bisector of $BC$ (so $OB = OC$).
Therefore,$OA = OB = OC$,which means point $O$ is the required point equidistant from all vertices $A, B,$ and $C$. This point $O$ is known as the circumcenter of the triangle.

(N/A) Consider a $\Delta ABC$.
Draw the angle bisector $'l'$ of $\angle B$.
Draw the angle bisector $'m'$ of $\angle C$.
Let the two angle bisectors $'l'$ and $'m'$ intersect at point $O$.
Since any point on the angle bisector of an angle is equidistant from the arms of the angle,point $O$ lies on the bisector of $\angle B$ (so it is equidistant from $AB$ and $BC$) and also on the bisector of $\angle C$ (so it is equidistant from $BC$ and $AC$).
Therefore,point $O$ is equidistant from all three sides ($AB$,$BC$,and $AC$) of $\Delta ABC$. This point $O$ is known as the incenter of the triangle.

(N/A) To ensure the ice cream parlour is equidistant from points $A$,$B$,and $C$,we need to find the circumcenter of the triangle formed by these points.
$1$. Join points $A$ and $B$,and draw line $l$,which is the perpendicular bisector of segment $AB$.
$2$. Join points $B$ and $C$,and draw line $m$,which is the perpendicular bisector of segment $BC$.
$3$. Let the perpendicular bisectors $l$ and $m$ intersect at point $O$.
Point $O$ is the circumcenter of $\triangle ABC$,and it is equidistant from vertices $A$,$B$,and $C$. Therefore,the ice cream parlour should be set up at point $O$.

(B) To find the number of equilateral triangles of side $1\,cm$ that can fit into the given shapes,we calculate the area of the shapes and divide by the area of one small equilateral triangle.
$1$. For the regular hexagon of side $s = 5\,cm$:
Area $= \frac{3\sqrt{3}}{2} s^2 = \frac{3\sqrt{3}}{2} \times 25 = 37.5\sqrt{3} \approx 64.95\,cm^2$.
Each small triangle of side $1\,cm$ has an area of $\frac{\sqrt{3}}{4} \times 1^2 = 0.25\sqrt{3} \approx 0.433\,cm^2$.
Number of triangles in Fig. $(i) = \frac{37.5\sqrt{3}}{0.25\sqrt{3}} = 150$.
$2$. For the star-shaped figure in Fig. $(ii)$:
This figure consists of a central hexagon of side $5\,cm$ and $6$ equilateral triangles of side $5\,cm$ attached to its sides.
Total area $= \text{Area of hexagon} + 6 \times \text{Area of triangle of side } 5\,cm$.
Total area $= 37.5\sqrt{3} + 6 \times (6.25\sqrt{3}) = 37.5\sqrt{3} + 37.5\sqrt{3} = 75\sqrt{3} \approx 129.9\,cm^2$.
Number of triangles in Fig. $(ii) = \frac{75\sqrt{3}}{0.25\sqrt{3}} = 300$.
Therefore,the star-shaped figure $(ii)$ has more triangles.