NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Question 3: $\small AD$ and $\small BC$ are equal perpendiculars to a line segment $\small AB$ (see Fig.). Show that $\small CD$ bisects $\small AB$ .

Answer:

In the given figure, consider $\Delta$ AOD and $\Delta$ BOC.

(i) AD = BC (given)

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

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

Thus, by the AAS Postulate, we have

$\Delta AOD\ \cong \ \Delta BOC$

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

And thus CD bisects AB.

Question 4: $\small l$ and $\small m$ are two parallel lines intersected by another pair of parallel lines $\small p$ and $\small q$ (see Fig. ). Show that $\small \Delta ABC\cong \Delta CDA$

Answer:

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

(i) $\angle\ BCA\ =\ \angle DAC$

(ii) $\angle\ BAC\ =\ \angle DCA$

(iii) Side AC is common in both triangles.

Thus, by ASA congruence, we have :

$\Delta ABC\ \cong \ \Delta CDA$

Question 5: (i) Line $\small l$ is the bisector of an angle $\small \angle A$ and B is any point on $\small l$ .$\small BP$ and $\small BQ$ are perpendiculars from $\small B$ to the arms of $\small \angle A$ (see Fig.). Show that: $\small \Delta APB\cong \Delta AQB$

Answer:

In the given figure consider $\small \Delta APB$ and $\small \Delta AQB$ ,

(i) $\angle P\ =\ \angle Q$ (Right angle)

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

(iii) Side AB is common in both triangles.

Thus, AAS congruence, we can write :

$\small \Delta APB\cong \Delta AQB$

Question 5: (ii) Line $\small l$ is the bisector of an angle $\small \angle A$ and $\small B$ is any point on $\small l$. $\small BP$ and $\small BQ$ are perpendiculars from $\small B$ to the arms of $\small \angle A$ (see Fig. ). Show that: $\small BP=BQ$ or $\small B$ is equidistant from the arms of $\small \angle A$ .

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.

Question 6: In Fig, $\small AC=AE,AB=AD$ and $\small \angle BAD= \angle EAC$ . Show that $\small BC=DE$.

Answer:

From the given figure, the following result can be drawn:-

$\angle BAD\ =\ \angle EAC$

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

$\angle BAD\ +\ \angle DAC\ =\ \angle EAC\ +\ \angle DAC$

$\angle BAC\ =\ \angle EAD$

Now consider $\Delta ABC$ and $\Delta ADE$ ,

(i) $AC\ =\ AE$ (Given)

(ii) $\angle BAC\ =\ \angle EAD$ (proved above)

(iii) $AB\ =\ AD$ (Given)

Thus, by SAS congruence, we can say that :

$\Delta ABC\ \cong \ \Delta ADE$

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

Question 7: (i) $\small AB$ is a line segment and $\small P$ is its mid-point. $\small D$ and $\small E$ are points on the same side of $\small AB$ such that $\small \angle BAD=\angle ABE$ and $\small \angle EPA=\angle DPB$ (see Fig). Show that $\small \Delta DAP\cong \Delta EBP$

Answer:

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

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

$\angle EPA\ +\ \angle DPE =\ \angle DPB\ +\ \angle DPE$

or $\angle DPA =\ \angle EPB$

Now, consider $\Delta DAP$ and $\Delta EBP$ :

(i) $\angle DPA =\ \angle EPB$ (Proved above)

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

(iii) $\small \angle BAD=\angle ABE$ (Given)

Hence, by ASA congruence, we can say that :

$\small \Delta DAP\cong \Delta EBP$

Question 7: (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 $\small \angle BAD=\angle ABE$ and $\small \angle EPA=\angle DPA$ (see Fig). Show that $\small AD=BE$

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 :

$\small AD=BE$

Question 8: (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 $\small DM=CM$. Point D is joined to point B (see Fig.). Show that: $\small \Delta AMC\cong \Delta BMD$

Answer:

Consider $\Delta AMC$ and $\Delta BMD$ ,

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

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

(iii) $CM\ =\ DM$ (Given)

Thus, by SAS congruency, we can conclude that :

$\small \Delta AMC\cong \Delta BMD$

Question 8: (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 $\small DM=CM$. Point D is joined to point B (see Fig.). Show that: $\small \angle DBC$ is a right angle.

Answer:

In the previous part, we have proved that $\small \Delta AMC\cong \Delta BMD$

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

This implies that side AC is parallel to BD.

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

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

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

Hence, $\small \angle DBC$ is a right angle.

Question 8: (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 $\small DM=CM$. Point D is joined to point B (see Fig.). Show that: $\small \Delta DBC\cong \Delta ACB$

Answer:

Consider $\Delta DBC$ and $\Delta ACB$ ,

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

(ii) $\angle ACB\ =\ \angle DBC$ (Right angle)

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

Thus, SAS congruence, we can conclude that :

$\small \Delta DBC\cong \Delta ACB$

Question 8: (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 $\small DM=CM$. Point D is joined to point B (see Fig.). Show that: $\small CM=\frac{1}{2}AB$

Answer:

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

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

$DM\ +\ CM\ =\ AM\ +\ BM$

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

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

Hence proved.

Question 1: (i) In an isosceles triangle ABC, with $\small AB=AC$, the bisectors of $\small \angle B$ and $\small \angle C$ intersect each other at O. Join A to O. Show that: $\small OB=OC$

Answer:

In the triangle ABC,

Since AB = AC, thus $\angle B\ =\ \angle C$

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

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

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

Question 1: (ii) In an isosceles triangle ABC, with $\small AB=AC$, the bisectors of $\small \angle B$ and $\small \angle C$ intersect each other at O. Join A to O. Show that: AO bisects $\small \angle A$

Answer:

Consider $\Delta AOB$ and $\Delta AOC$ ,

(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 :

$\Delta AOB\ \cong \ \Delta AOC$

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

$\angle BAO\ =\ \angle CAO$

Hence, AO bisects $\angle A$.

Question 2: In $\Delta ABC$ AD is the perpendicular bisector of BC (see Fig). Show that $\small \Delta ABC$ is an isosceles triangle in which $\small AB=AC$ .

Answer:

Consider $\Delta$ ABD and $\Delta$ ADC,

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

(ii) $\angle ADB\ =\ \angle ADC$ (Right angle)

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

Thus, by the SAS congruence axiom, we can state :

$\Delta ADB\ \cong \ \Delta ADC$

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

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

Question 3: 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 $\Delta AEB$ and $\Delta AFC$ ,

(i) $\angle A$ is common to both triangles.

(ii) $\angle AEB\ =\ \angle AFC$ (Right angles)

(iii) $AB\ =\ AC$ (Given)

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

$\Delta AEB\ \cong \Delta AFC$

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

Hence, these altitudes are equal.

Question 4: (i) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that $\small \Delta ABE \cong \Delta ACF$

Answer:

Consider $\Delta ABE$ and $\Delta ACF$ ,

(i) $\angle A$ is common to both triangles.

(ii) $\angle AEB\ =\ \angle AFC$ (Right angles)

(iii) $BE\ =\ CF$ (Given)

Thus, by AAS congruence, we can say that :

$\small \Delta ABE \cong \Delta ACF$

Question 4: (ii) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig.). Show that $\small AB=AC$, i.e., ABC is an isosceles triangle.

Answer:

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

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

Hence, $\Delta \ ABC$ is an isosceles triangle.

Question 5: ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that $\small \angle ABD\ \cong \ \angle ACD$.

Answer:

Consider $\Delta ABD$ and $\Delta ACD$ ,

(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 :

$\small \angle ABD\ \cong \ \angle ACD$

Question 6: $\Delta ABC$ is an isosceles triangle in which $AB=AC$. Side BA is produced to D such that $AD=AB$ (see Fig.). Show that $\angle BCD$ is a right angle.

Answer:

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

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

Similarly in $\Delta$ ACD,

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

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

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

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

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

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

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

Question 7: ABC is a right-angled triangle in which $\small \angle A =90^{\circ}$ and $\small AB=AC$. Find $\small \angle B$ and $\small \angle C$.

Answer:

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

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

Thus, $\angle B\ =\ \angle C$

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

So, we have :

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

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

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

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

Question 8: Show that the angles of an equilateral triangle are $\small 60^{\circ}$ 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 : $\angle A\ =\ \angle B\ =\ \angle C$

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

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

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

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

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

Question 1: (i) $\small \Delta ABC$ and $\small \Delta DBC$ 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 $\small \Delta ABD\cong \Delta ACD$

Answer:

Consider $\Delta ABD$ and $\Delta ACD$ ,

(i) $AD\ =\ AD$ (Common)

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

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

Thus, by SSS congruency, we can conclude that :

$\small \Delta ABD\cong \Delta ACD$

Question 1: (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 $\small \Delta ABP \cong \Delta ACP.$

Answer:

Consider $\Delta ABP$ and $\Delta ACP$,

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

(ii) $\angle PAB\ =\ \angle PAC$ (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 :

$\small \Delta ABP \cong \Delta ACP$

Question 1: (iii) $\small \Delta ABC$ and $\small \Delta DBC$ 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 $\small \angle A$ as well as $\small \angle D$.

Answer:

In the first part, we have proved that $\small \Delta ABD\cong \Delta ACD$

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

Hence, AP bisects $\angle A$.

Now consider $\Delta BPD$ and $\Delta CPD$ ,

(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 :

$\Delta BPD\ \cong \ \Delta CPD$

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

or AP bisects $\angle D$ .

Question 1: (iv) $\small \Delta ABC$ and $\small \Delta DBC$ 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 : $\angle BPD\ =\ \angle CPD$

Also, $BP\ =\ CP$

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

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

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

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

Question 2: (i) AD is an altitude of an isosceles triangle ABC in which $\small AB=AC$. Show that AD bisects BC

Answer:

Consider $\Delta ABD$ and $\Delta ACD$ ,

(i) $AB\ =\ AC$ (Given)

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

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

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

$\Delta ABD\ \cong \ \Delta ACD$

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

Question 2: (ii) AD is an altitude of an isosceles triangle ABC in which $\small AB=AC$. Show that AD bisects $\small \angle A$

Answer:

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

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

$\angle BAD\ =\ \angle CAD$

Hence, $AD$ bisects $\angle A$ .

Question 3: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $\small \Delta PQR$ (see Fig). Show that:

(i) $\small \Delta ABM \cong \Delta PQN$

(ii) $\small \Delta ABC \cong \Delta PQR$

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 $\Delta ABM$ and $\Delta PQN$ ,

(a) $AM\ =\ PN$ (Given)

(b) $AB\ =\ PQ$ (Given)

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

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

$\small \Delta ABM \cong \Delta PQN$

(ii) Consider $\Delta ABC$ and $\Delta PQR$ :

(a) $AB\ =\ PQ$ (Given)

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

(c) $BC\ =\ QR$ (Given)

Thus, by the SAS congruence rule,

$\small \Delta ABC \cong \Delta PQR$

Question 4: 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 $\Delta BEC$ and $\Delta CFB$ ,

(i) $\angle BEC\ =\ \angle CFB$ (Right angle)

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

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

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

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

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

Question 5: ABC is an isosceles triangle with $\small AB=AC$. Draw $\small AP\perp BC$ to show that $\small \angle B=\angle C$ .

Answer:

Consider $\Delta ABP$ and $\Delta ACP$ ,

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

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

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

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

$\Delta ABP\ \cong \Delta ACP$

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

$\angle B\ =\ \angle C$