NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Q1 In quadrilateral $ABCD$ , $AC=AD$ and $AB$ bisects $\mathrm{\angle }A$ (see Fig.). Show that $\mathrm{\Delta }ABC\cong \mathrm{\Delta }ABD$ . What can you say about $BC$ and $BD$?

Answer:

In the given triangles we are given that:-

(i) $AC=AD$

(ii) Further, it is given that AB bisects angle A. Thus $\mathrm{\angle }$ BAC $=\mathrm{\angle }$ BAD.

(iii) Side AB is common in both the triangles. $AB=AB$

Hence by SAS congruence, we can say that: $\mathrm{\Delta }ABC\cong \mathrm{\Delta }ABD$

By c.p.c.t. (corresponding parts of congruent triangles are equal) we can say that $BC=BD$

Q2 (i) $ABCD$ is a quadrilateral in which $AD=BC$ and $\mathrm{\angle }DAB=\mathrm{\angle }CBA$ (see Fig. ). Prove that $\mathrm{\Delta }ABD\cong \mathrm{\Delta }BAC$ .

Answer:

It is given that:-

(i) AD = BC

(ii) $\mathrm{\angle }DAB=\mathrm{\angle }CBA$

(iii) Side AB is common in both the triangles.

So, by SAS congruence, we can write :

$\mathrm{\Delta }ABD\cong \mathrm{\Delta }BAC$

Q2 (ii) $ABCD$ is a quadrilateral in which $AD=BC$ and $\mathrm{\angle }DAB=\mathrm{\angle }CBA$ (see Fig.). Prove that $BD=AC$

Answer:

In the previous part, we have proved that $\mathrm{\Delta }ABD\cong \mathrm{\Delta }BAC$.

Thus by c.p.c.t. , we can write : $BD=AC$

Q2 (iii) $ABCD$ is a quadrilateral in which $AD=BC$ and $\mathrm{\angle }DAB=\mathrm{\angle }BAC$ (see Fig.). Prove that $\mathrm{\angle }ABD=\mathrm{\angle }BAC$ .

Answer:

In the first part, we have proved that $\mathrm{\Delta }ABD\cong \mathrm{\Delta }BAC$.

Thus by c.p.c.t. , we can conclude :

$\mathrm{\angle }ABD=\mathrm{\angle }BAC$

Q3 $AD$ and $BC$ are equal perpendiculars to a line segment $AB$ (see Fig.). Show that $CD$ bisects $AB$ .

Answer:

In the given figure consider $\mathrm{\Delta }$ AOD and $\mathrm{\Delta }$ BOC.

(i) AD = BC (given)

(ii) $\mathrm{\angle }$ A = $\mathrm{\angle }$ B (given that the line AB is perpendicular to AD and BC)

(iii) $\mathrm{\angle }$ AOD = $\mathrm{\angle }$ BOC (vertically opposite angles).

Thus by AAS Postulate, we have

$\mathrm{\Delta }AOD\cong \mathrm{\Delta }BOC$

Hence by c.p.c.t. we can write : $AO=OB$

And thus CD bisects AB.

Q4 $l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see Fig. ). Show that $\mathrm{\Delta }ABC\cong \mathrm{\Delta }CDA$

Answer:

In the given figure, consider $\mathrm{\Delta }$ ABC and $\mathrm{\Delta }$ CDA :

(i) $\mathrm{\angle }BCA=\mathrm{\angle }DAC$

(ii) $\mathrm{\angle }BAC=\mathrm{\angle }DCA$

(iii) Side AC is common in both the triangles.

Thus by ASA congruence, we have :

$\mathrm{\Delta }ABC\cong \mathrm{\Delta }CDA$

Q5 (i) Line $l$ is the bisector of an angle $\mathrm{\angle }A$ and B is any point on $l$ .$BP$ and $BQ$ are perpendiculars from $B$ to the arms of $\mathrm{\angle }A$ (see Fig.). Show that: $\mathrm{\Delta }APB\cong \mathrm{\Delta }AQB$

Answer:

In the given figure consider $\mathrm{\Delta }APB$ and $\mathrm{\Delta }AQB$ ,

(i) $\mathrm{\angle }P=\mathrm{\angle }Q$ (Right angle)

(ii) $\mathrm{\angle }BAP=\mathrm{\angle }BAQ$ (Since it is given that I is bisector)

(iii) Side AB is common in both triangles.

Thus, AAS congruence, we can write :

$\mathrm{\Delta }APB\cong \mathrm{\Delta }AQB$

Q5 (ii) Line $l$ is the bisector of an angle $\mathrm{\angle }A$ and $B$ is any point on $l$. $BP$ and $BQ$ are perpendiculars from $B$ to the arms of $\mathrm{\angle }A$ (see Fig. ). Show that: $BP=BQ$ or $B$ is equidistant from the arms of $\mathrm{\angle }A$ .

Answer:

In the previous part, we have proved that $\small \Delta APB\cong \Delta AQB.

Thus, by c.p.c.t.. we can write :

$BP=BQ$

Thus B, is equidistant from the arms of angle A.

Q6 In Fig, $AC=AE,AB=AD$ and $\mathrm{\angle }BAD=\mathrm{\angle }EAC$ . Show that $BC=DE$.

Answer:

From the given figure following result can be drawn:-

$\mathrm{\angle }BAD=\mathrm{\angle }EAC$

Adding $\mathrm{\angle }DAC$ to both sides, we get :

$\mathrm{\angle }BAD+\mathrm{\angle }DAC=\mathrm{\angle }EAC+\mathrm{\angle }DAC$

$\mathrm{\angle }BAC=\mathrm{\angle }EAD$

Now consider $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }ADE$ , :-

(i) $AC=AE$ (Given)

(ii) $\mathrm{\angle }BAC=\mathrm{\angle }EAD$ (proved above)

(iii) $AB=AD$ (Given)

Thus by SAS congruence, we can say that :

$\mathrm{\Delta }ABC\cong \mathrm{\Delta }ADE$

Hence by c.p.c.t., we can say that : $BC=DE$

Q7 (i) $AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\mathrm{\angle }BAD=\mathrm{\angle }ABE$ and $\mathrm{\angle }EPA=\mathrm{\angle }DPB$ (see Fig). Show that $\mathrm{\Delta }DAP\cong \mathrm{\Delta }EBP$

Answer:

From the figure, it is clear that :
$\mathrm{\angle }EPA=\mathrm{\angle }DPB$

Adding $\mathrm{\angle }DPE$ on both sides, we get :

$\mathrm{\angle }EPA+\mathrm{\angle }DPE=\mathrm{\angle }DPB+\mathrm{\angle }DPE$

or $\mathrm{\angle }DPA=\mathrm{\angle }EPB$

Now, consider $\mathrm{\Delta }DAP$ and $\mathrm{\Delta }EBP$ :

(i) $\mathrm{\angle }DPA=\mathrm{\angle }EPB$ (Proved above)

(ii) $AP=BP$ (Since P is the midpoint of line AB)

(iii) $\mathrm{\angle }BAD=\mathrm{\angle }ABE$ (Given)

Hence by ASA congruence, we can say that :

$\mathrm{\Delta }DAP\cong \mathrm{\Delta }EBP$

Q7 (ii) AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that $\mathrm{\angle }BAD=\mathrm{\angle }ABE$ and $\mathrm{\angle }EPA=\mathrm{\angle }DPA$ (see Fig). Show that $AD=BE$

Answer:

In the previous part, we have proved that $\small \Delta DAP\cong \Delta EBP.

Thus by c.p.c.t., we can say that :

$AD=BE$

Q8 (i) In right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that $DM=CM$. Point D is joined to point B (see Fig.). Show that: $\mathrm{\Delta }AMC\cong \mathrm{\Delta }BMD$

Answer:

Consider $\mathrm{\Delta }AMC$ and $\mathrm{\Delta }BMD$ ,

(i) $AM=BM$ (Since M is the mid-point)

(ii) $\mathrm{\angle }CMA=\mathrm{\angle }DMB$ (Vertically opposite angles are equal)

(iii) $CM=DM$ (Given)

Thus by SAS congruency, we can conclude that :

$\mathrm{\Delta }AMC\cong \mathrm{\Delta }BMD$

Q8 (ii) In right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that $DM=CM$. Point D is joined to point B (see Fig.). Show that: $\mathrm{\angle }DBC$ is a right angle.

Answer:

In the previous part, we have proved that $\mathrm{\Delta }AMC\cong \mathrm{\Delta }BMD$

By c.p.c.t. we can say that : $\mathrm{\angle }ACM=\mathrm{\angle }BDM$

This implies side AC is parallel to BD.

Thus we can write : $\mathrm{\angle }ACB+\mathrm{\angle }DBC={180}^{\circ }$ (Co-interior angles)

and, ${90}^{\circ }+\mathrm{\angle }DBC={180}^{\circ }$

or $\mathrm{\angle }DBC={90}^{\circ }$

Hence, $\mathrm{\angle }DBC$ is a right angle.

Q8 (iii) In right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that $DM=CM$. Point D is joined to point B (see Fig.). Show that: $\mathrm{\Delta }DBC\cong \mathrm{\Delta }ACB$

Answer:

Consider $\mathrm{\Delta }DBC$ and $\mathrm{\Delta }ACB$ ,

(i) $BC=BC$ (Common in both the triangles)

(ii) $\mathrm{\angle }ACB=\mathrm{\angle }DBC$ (Right angle)

(iii) $DB=AC$ (By c.p.c.t. from the part (a) of the question.)

Thus, SAS congruence, we can conclude that :

$\mathrm{\Delta }DBC\cong \mathrm{\Delta }ACB$

Q8 (iv) In right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that $DM=CM$. Point D is joined to point B (see Fig.). Show that: $CM=\frac{1}{2}AB$

Answer:

In the previous part, we have proved that $\mathrm{\Delta }DBC\cong \mathrm{\Delta }ACB$.

Thus by c.p.c.t., we can write : $DC=AB$

$DM+CM=AM+BM$

or $CM+CM=AB$ (Since M is midpoint.)

or $CM=\frac{1}{2}AB$ .

Hence proved.

Q1 (i) In an isosceles triangle ABC, with $AB=AC$, the bisectors of $\mathrm{\angle }B$ and $\mathrm{\angle }C$ intersect each other at O. Join A to O. Show that: $OB=OC$

Answer:

In the triangle ABC,

Since AB = AC, thus $\mathrm{\angle }B=\mathrm{\angle }C$

or $\frac{1}{2}\mathrm{\angle }B=\frac{1}{2}\mathrm{\angle }C$

or $\mathrm{\angle }OBC=\mathrm{\angle }OCB$ (Angles bisectors are equal)

Thus $OB=OC$ as sides opposite to equal are angles are also equal.

Q1 (ii) In an isosceles triangle ABC, with $AB=AC$, the bisectors of $\mathrm{\angle }B$ and $\mathrm{\angle }C$ intersect each other at O. Join A to O. Show that: AO bisects $\mathrm{\angle }A$

Answer:

Consider $\mathrm{\Delta }AOB$ and $\mathrm{\Delta }AOC$ ,

(i) $AB=AC$ (Given)

(ii) $AO=AO$ (Common in both the triangles)

(iii) $OB=OC$ (Proved in previous part)

Thus by the SSS congruence rule, we can conclude that :

$\mathrm{\Delta }AOB\cong \mathrm{\Delta }AOC$

Now, by c.p.c.t.,

$\mathrm{\angle }BAO=\mathrm{\angle }CAO$

Hence AO bisects $\mathrm{\angle }A$.

Q2 In $\mathrm{\Delta }ABC$ AD is the perpendicular bisector of BC (see Fig). Show that $\mathrm{\Delta }ABC$ is an isosceles triangle in which $AB=AC$ .

Answer:

Consider $\mathrm{\Delta }$ ABD and $\mathrm{\Delta }$ ADC,

(i) $AD=AD$ (Common in both the triangles)

(ii) $\mathrm{\angle }ADB=\mathrm{\angle }ADC$ (Right angle)

(iii) $BD=CD$ (Since AD is the bisector of BC)

Thus, by SASthe congruence axiom, we can state :

$\mathrm{\Delta }ADB\cong \mathrm{\Delta }ADC$

Hence, by c.p.c.t., we can say that : $AB=AC$

Thus $\mathrm{\Delta }ABC$ is an isosceles triangle with AB and AC as equal sides.

Q3 ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig). Show that these altitudes are equal.

Answer:

Consider $\mathrm{\Delta }AEB$ and $\mathrm{\Delta }AFC$ ,

(i) $\mathrm{\angle }A$ is common in both the triangles.

(ii) $\mathrm{\angle }AEB=\mathrm{\angle }AFC$ (Right angles)

(iii) $AB=AC$ (Given)

Thus, by the AAS congruence axiom, we can conclude that :

$\mathrm{\Delta }AEB\cong \mathrm{\Delta }AFC$

Now, by c.p.c.t. we can say : $BE=CF$

Hence, these altitudes are equal.

Q4 (i) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that $\mathrm{\Delta }ABE\cong \mathrm{\Delta }ACF$

Answer:

Consider $\mathrm{\Delta }ABE$ and $\mathrm{\Delta }ACF$ ,

(i) $\mathrm{\angle }A$ is common in both the triangles.

(ii) $\mathrm{\angle }AEB=\mathrm{\angle }AFC$ (Right angles)

(iii) $BE=CF$ (Given)

Thus by AAS congruence, we can say that :

$\mathrm{\Delta }ABE\cong \mathrm{\Delta }ACF$

Q4 (ii) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig.). Show that $AB=AC$, i.e., ABC is an isosceles triangle.

Answer:

From the previous part of the question we found out that: $\mathrm{\Delta }ABE\cong \mathrm{\Delta }ACF$

Now, by c.p.c.t. we can say that : $AB=AC$

Hence, $\mathrm{\Delta }ABC$ is an isosceles triangle.

Q5. ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that $\mathrm{\angle }ABD\cong \mathrm{\angle }ACD$.

Answer:

Consider $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACD$ ,

(i) $AD=AD$ (Common in both the triangles)

(ii) $AB=AC$ (Sides of isosceles triangle)

(iii) $BD=CD$ (Sides of isosceles triangle)

Thus, by SSS congruency, we can conclude that :

$\mathrm{\angle }ABD\cong \mathrm{\angle }ACD$

Q6 $\mathrm{\Delta }ABC$ is an isosceles triangle in which $AB=AC$. Side BA is produced to D such that $AD=AB$ (see Fig.). Show that $\mathrm{\angle }BCD$ is a right angle.

Answer:

Consider $\mathrm{\Delta }$ ABC,
It is given that AB = AC

So, $\mathrm{\angle }ACB=\mathrm{\angle }ABC$ (Since angles opposite to the equal sides are equal.)

Similarly in $\mathrm{\Delta }$ ACD,

We have AD = AB
and $\mathrm{\angle }ADC=\mathrm{\angle }ACD$
So,

$\mathrm{\angle }CAB+\mathrm{\angle }ACB+\mathrm{\angle }ABC={180}^{\circ }$

$\mathrm{\angle }CAB+2\mathrm{\angle }ACB={180}^{\circ }$
or $\mathrm{\angle }CAB={180}^{\circ }-2\mathrm{\angle }ACB$ ...........................(i)

And in $\mathrm{\Delta }$ ADC,
$\mathrm{\angle }CAD={180}^{\circ }-2\mathrm{\angle }ACD$ ..............................(ii)

Adding (i) and (ii), we get :
$\mathrm{\angle }CAB+\mathrm{\angle }CAD={360}^{\circ }-2\mathrm{\angle }ACD-2\mathrm{\angle }ACB$

or ${180}^{\circ }={360}^{\circ }-2\mathrm{\angle }ACD-2\mathrm{\angle }ACB$

and $\mathrm{\angle }BCD={90}^{\circ }$

Q7 ABC is a right-angled triangle in which $\mathrm{\angle }A={90}^{\circ }$ and $AB=AC$. Find $\mathrm{\angle }B$ and $\mathrm{\angle }C$.

Answer:

In the triangle ABC, sides AB and AC are equal.

We know that angles opposite to equal sides are also equal.

Thus, $\mathrm{\angle }B=\mathrm{\angle }C$

Also, the sum of the interior angles of a triangle is ${180}^{\circ }$

So, we have :

$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$

or ${90}^{\circ }+2\mathrm{\angle }B={180}^{\circ }$

or $\mathrm{\angle }B={45}^{\circ }$

Hence, $\mathrm{\angle }B=\mathrm{\angle }C={45}^{\circ }$

Q8 Show that the angles of an equilateral triangle are ${60}^{\circ }$ each.

Answer:

Consider a triangle ABC that has all sides equal.

We know that angles opposite to equal sides are equal.

Thus we can write : $\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C$

Also, the sum of the interior angles of a triangle is ${180}^{\circ }$ .

Hence, $\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$

or $3\mathrm{\angle }A={180}^{\circ }$

or $\mathrm{\angle }A={60}^{\circ }$

So, all the angles of the equilateral triangle are equal ( ${60}^{\circ }$ ).

Q1 (i) $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that $\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$

Answer:

Consider $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACD$ ,

(i) $AD=AD$ (Common)

(ii) $AB=AC$ (Isosceles triangle)

(iii) $BD=CD$ (Isosceles triangle)

Thus by SS,S congruency, we can conclude that :

$\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$

Q1 (ii) Triangles ABC and Triangle DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig). If AD is extended to intersect BC at P, show that $\mathrm{\Delta }ABP\cong \mathrm{\Delta }ACP.$

Answer:

Consider $\mathrm{\Delta }ABP$ and $\mathrm{\Delta }ACP$,

(i) $AP$ is the common side in both the triangles.

(ii) $\mathrm{\angle }PAB=\mathrm{\angle }PAC$ (This is obtained from the c.p.c.t. as proved in the previous part.)

(iii) $AB=AC$ (Isosceles triangles)

Thus, by the SAS axiom, we can conclude that :

$\mathrm{\Delta }ABP\cong \mathrm{\Delta }ACP$

Q1 (iii) $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that AP bisects $\mathrm{\angle }A$ as well as $\mathrm{\angle }D$.

Answer:

In the first part, we have proved that $\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$

So, by c.p.c.t. $\mathrm{\angle }PAB=\mathrm{\angle }PAC$ .

Hence AP bisects $\mathrm{\angle }A$.

Now consider $\mathrm{\Delta }BPD$ and $\mathrm{\Delta }CPD$ ,

(i) $PD=PD$ (Common)

(ii) $BD=CD$ (Isosceles triangle)

(iii) $BP=CP$ (by c.p.c.t. from the part (b))

Thus, by SSS congruency, we have :

$\mathrm{\Delta }BPD\cong \mathrm{\Delta }CPD$

Hence by c.p.c.t. we have : $\mathrm{\angle }BDP=\mathrm{\angle }CDP$

or AP bisects $\mathrm{\angle }D$ .

Q1 (iv) $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that AP is the perpendicular bisector of BC.

Answer:

In the previous part, we have proved that $\Delta BPD\ \cong \ \Delta CPD.

Thus by c.p.c.t. we can say that : $\mathrm{\angle }BPD=\mathrm{\angle }CPD$

Also, $BP=CP$

Since BC is a straight line, thus : $\mathrm{\angle }BPD+\mathrm{\angle }CPD={180}^{\circ }$

or $2\mathrm{\angle }BPD={180}^{\circ }$

or $\mathrm{\angle }BPD={90}^{\circ }$

Hence, it is clear that AP is a perpendicular bisector of line BC.

Q2 (i) AD is an altitude of an isosceles triangle ABC in which $AB=AC$. Show that AD bisects BC

Answer:

Consider $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACD$ ,

(i) $AB=AC$ (Given)

(ii) $AD=AD$ (Common in both triangles)

(iii) $\mathrm{\angle }ADB=\mathrm{\angle }ADC={90}^{\circ }$

Thus, by the RHS axiom, we can conclude that :

$\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$

Hence, by c.p.c.t. we can say that: $BD=CD$ or AD bisects BC.

Q2 (ii) AD is an altitude of an isosceles triangle ABC in which $AB=AC$. Show that AD bisects $\mathrm{\angle }A$

Answer:

In the previous part of the question, we have proved that $\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$

Thus by c.p.c.t., we can write :

$\mathrm{\angle }BAD=\mathrm{\angle }CAD$

Hence, $AD$ bisects $\mathrm{\angle }A$ .

Q3 Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $\mathrm{\Delta }PQR$ (see Fig). Show that:

(i) $\mathrm{\Delta }ABM\cong \mathrm{\Delta }PQN$

(ii) $\mathrm{\Delta }ABC\cong \mathrm{\Delta }PQR$

Answer:

(i) From the figure we can say that :

$BC=QR$

or $\frac{1}{2}BC=\frac{1}{2}QR$

or $BM=QN$

Now, consider $\mathrm{\Delta }ABM$ and $\mathrm{\Delta }PQN$ ,

(a) $AM=PN$ (Given)

(b) $AB=PQ$ (Given)

(c) $BM=QN$ (Prove above)

Thus, by the SSS congruence rule, we can conclude that :

$\mathrm{\Delta }ABM\cong \mathrm{\Delta }PQN$

(ii) Consider $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }PQR$ :

(a) $AB=PQ$ (Given)

(b) $\mathrm{\angle }ABC=\mathrm{\angle }PQR$ (by c.p.c.t. from the above proof)

(c) $BC=QR$ (Given)

Thus, by the SAS congruence rule,

$\mathrm{\Delta }ABC\cong \mathrm{\Delta }PQR$

Q4 BE and CF are two equal altitudes of a triangle ABC. Using the RHS congruence rule, prove that the triangle ABC is isosceles.

Answer:

Using the given conditions, consider $\mathrm{\Delta }BEC$ and $\mathrm{\Delta }CFB$ ,

(i) $\mathrm{\angle }BEC=\mathrm{\angle }CFB$ (Right angle)

(ii) $BC=BC$ (Common in both the triangles)

(iii) $BE=CF$ (Given that altitudes are of the same length. )

Thus by RHS axiom, we can say that: $\mathrm{\Delta }BEC\cong \mathrm{\Delta }CFB$

Hence by c.p.c.t., $\mathrm{\angle }B=\mathrm{\angle }C$

And thus $AB=AC$ (sides opposite to equal angles are also equal). Thus ABC is an isosceles triangle.

Q5 ABC is an isosceles triangle with $AB=AC$. Draw $AP\perp BC$ to show that $\mathrm{\angle }B=\mathrm{\angle }C$ .

Answer:

Consider $\mathrm{\Delta }ABP$ and $\mathrm{\Delta }ACP$ ,

(i) $\mathrm{\angle }APB=\mathrm{\angle }APC={90}^{\circ }$ (Since it is given that AP is altitude.)

(ii) $AB=AC$ (Isosceles triangle)

(iii) $AP=AP$ (Common in both triangles)

Thus, by the RHS axiom, we can conclude that :

$\mathrm{\Delta }ABP\cong \mathrm{\Delta }ACP$

Now, by c.p.c.t. we can say that :

$\mathrm{\angle }B=\mathrm{\angle }C$