# NCERT Solutions for Class 9 Maths Chapter 7 Triangles  Written by Team Trustudies
Updated at 2021-02-14

## NCERT solutions for class 9 Maths Chapter 7 Triangles Exercise 7.1

Q1 ) In quadrilateral ACBD, AC = AD and AB bisects $\mathrm{?}A$ (see figure). Show that $\mathrm{?}ABC$ $?$ $\mathrm{?}ABD$. What can you say about BC and BD ? NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In $\mathrm{?}ABC$ and $\mathrm{?}ABD$, we have,

As, AB bisects $\mathrm{?}A$,

Therefore, $\mathrm{?}CAB$ = $\mathrm{?}DAB$

Also, AB is a common side,

So, we can say,
$\mathrm{?}ABC$ $?$ $\mathrm{?}ABD$ ...(by SAS test of congruency)

Also, BC = BD ...(By CPCT)
Hence, proved.

Q2 ) ABCD is a quadrilateral in which AD = BC and $\mathrm{?}DAB$ = $\mathrm{?}CBA$ (see figure). Prove that
(i) $\mathrm{?}ABD$ $?$ $\mathrm{?}BAC$
(ii) BD = AC
(iii) $\mathrm{?}ABD$ = $\mathrm{?}BAC$ NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In $\mathrm{?}ABC$ and $\mathrm{?}BAC$, we have

Also,
$\mathrm{?}DAB$ = $\mathrm{?}CBA$ ...(Given)

And AB is a common side.

Therefore, $\mathrm{?}ABD$ $?$ $\mathrm{?}BAC$ ...(SAS congruency test)

Hence, BD = AC ...(By CPCT)

and $\mathrm{?}ABD$ = $\mathrm{?}BAC$ ...(By CPCT)
Hence, proved.

Q3 ) AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB. NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In $\mathrm{?}AOD$ and $\mathrm{?}BOC$, we have,

Now, $\mathrm{?}AOD$ = $\mathrm{?}BOC$ ...(vertically opposite angles)

Also, $\mathrm{?}DAO$ = $\mathrm{?}CBO$ = ${90}^{?}$

and BD = BC ...(shown in the figure)

Therefore, $\mathrm{?}AOD$ $?$ $\mathrm{?}BOC$ ...(SAS congruency test)

Hence, OA = OB ...(By CPCT)

Thus, we can say that, O is the mid-point of AB.

So, CD bisects AB.
Hence, proved

Q4 ) l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that $\mathrm{?}ABC$ $?$ $\mathrm{?}CDA$. NCERT Solutions for Class 9 Maths Chapter 7 Triangles From figure, we have,

$\mathrm{?}1$ = $\mathrm{?}2$ (Vertically opposite angles) ...(i)

$\mathrm{?}1$ = $\mathrm{?}6$ (Corresponding angles) ...(ii)
$\mathrm{?}6$ = $\mathrm{?}4$ (Corresponding angles) ...(iii)

From Equations, (i), (ii) and (iii), we have

$\mathrm{?}1$ = $\mathrm{?}4$ and $\mathrm{?}2$ = $\mathrm{?}6$ ...(iv)

In $\mathrm{?}ABC$ and $\mathrm{?}CDA$, we have

$\mathrm{?}4$ = $\mathrm{?}2$ ...(from (iii) and (iv))
$\mathrm{?}5$ = $\mathrm{?}3$ ...(Alternate angles)

and AC is a common side.

Therefore, we get,
$\mathrm{?}ABC$ $?$ $\mathrm{?}CDA$ ...(By SAS congruency test)
Hence, proved.

Q5 ) Line l is the bisector of a $\mathrm{?}A$ and $\mathrm{?}B$ is any point on l. BP and BQ are perpendiculars from B to the arms of $\mathrm{?}A$ (see figure). show that
(i) $\mathrm{?}APB$ $?$ $\mathrm{?}AQB$
(ii) BP = BQ or B is equidistant from the arms of $\mathrm{?}A$. NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In $\mathrm{?}APB$ $?$ $\mathrm{?}AQB$, we have

$\mathrm{?}APB$ = $\mathrm{?}AQB$ = ${90}^{?}$
$\mathrm{?}PAB$ = $\mathrm{?}QAB$ ...(AB bisects $\mathrm{?}PAQ$)

AB is the common side.
$?$ $\mathrm{?}$ {APB}\) $?$ $\mathrm{?}AQB$ ...(By AAS congruency test)

Also, BP = BQ ...(By CPCT)

Thus, we can say that, B is equidistant from the arms of $\mathrm{?}A$.
Hence, proved.

Q6 ) In figure, AC = AE, AB = AD and $\mathrm{?}BAD$ = $\mathrm{?}EAC$. Show that BC = DE. NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In $\mathrm{?}ABC$ and $\mathrm{?}ADE$ , we have,

$\mathrm{?}BAD$ = $\mathrm{?}EAC$ ...(i)(Given)
On adding, $\mathrm{?}DAC$ on both sides in Eq. (i), we get,

$?$ $\mathrm{?}BAD$ + $\mathrm{?}DAC$ = $\mathrm{?}EAC$ + $\mathrm{?}DAC$
$?$ $\mathrm{?}BAC$ = $\mathrm{?}DAE$

and also, AC = AE ...(Given)

$?$ $\mathrm{?}ABC$ $?$ $\mathrm{?}ADE$ ...(By AAS congruency test)

Thus, BC = DE ...(By CPCT)

Hence, proved.

Q7 ) AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that $\mathrm{?}BAD$ = $\mathrm{?}ABE$ and $\mathrm{?}EPA$ = $\mathrm{?}DPB$(see figure). Show that
(i) $\mathrm{?}DAP$ $?$ $\mathrm{?}EBP$ NCERT Solutions for Class 9 Maths Chapter 7 Triangles

We have,
AP = BP ...(i)(Since, P is the mid-point of AB)

$\mathrm{?}EPA$ = $\mathrm{?}DPB$ ...(ii)(Given)
$\mathrm{?}BAD$ = $\mathrm{?}ABE$ ...(iii) (Given)
On adding, $\mathrm{?}EPD$ on both sides in Equation (ii),
we have,
$?$ $\mathrm{?}EPA$ + $\mathrm{?}EPD$ = $\mathrm{?}DPB$ + $\mathrm{?}EPD$
$?$ $\mathrm{?}DPA$ = $\mathrm{?}EPB$ ...(iv)

Now, In $\mathrm{?}DAP$ and $\mathrm{?}EBP$ We have,
$?$ $\mathrm{?}DPA$ = $\mathrm{?}EPB$ ...(from (iv)),
$?$ $\mathrm{?}DAP$ = $\mathrm{?}EBP$ ...(Given)
and AP = BP ...(From Eq. (i))

$?$ $\mathrm{?}DAP$ $?$ $\mathrm{?}EBP$ ...(By ASA congruency test)

Thus, AD = BE ...(By CPCT)

Hence, proved.

Q8 ) 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 figure). Show that
(i) $\mathrm{?}AMC$ $?$ $\mathrm{?}BMD$
(ii) $\mathrm{?}DBC$ is a right angle
(iii) $\mathrm{?}DBC$ $?$ $\mathrm{?}ACB$
(iv)CM = (1/2) AB

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Given :
$\mathrm{?}ACB$ in which $\mathrm{?}C$ = ${90}^{?}$ and M is the mid-point of AB.

To prove :
(i) $\mathrm{?}AMC$ $?$ $\mathrm{?}BMD$
(ii) $\mathrm{?}DBC$ is a right angle
(iii) $\mathrm{?}DBC$ $?$ $\mathrm{?}ACB$
(iv)CM = ($\frac{1}{2}$ ) AB Construction : Produce CM to D, such that CM = MD. Join DB.

Proof : In $\mathrm{?}AMC$ and $\mathrm{?}BMD$, we have

AM = BM ...(M is the mid-point of AB)
CM = DM ...(Given)
and $\mathrm{?}AMC$ = $\mathrm{?}BMD$ ...(Vertically opposite angles)
$?$ $\mathrm{?}AMC$ $?$ $\mathrm{?}BMD$(By SAS congruency test)
Hence, part(i) is proved.

Also, AC = DB ...(By CPCT)

$\mathrm{?}1$ = $\mathrm{?}2$ ...(Alternate angles) and (by CPCT)

$?$ BD || CA and BC is transversal.

$?$ $\mathrm{?}ACB$ + $\mathrm{?}DBC$ = ${180}^{?}$
But, $\mathrm{?}ACB$ = ${90}^{?}$ ...(given)
$?$ ${90}^{?}$ + $\mathrm{?}DBC$ = ${180}^{?}$
$?$ $\mathrm{?}DBC$ = ${180}^{?}$ - ${90}^{?}$
$?$ $\mathrm{?}DBC$ = ${90}^{?}$

Hence, part(ii) is proved, too.

Now, considering $\mathrm{?}DBC$ and $\mathrm{?}ACB$, we have,

AC = DB ...(from part(i))

Side BC is common.

and $\mathrm{?}DBC$ = $\mathrm{?}ACB$ = ${90}^{?}$

Therefore, $\mathrm{?}DBC$ $?$ $\mathrm{?}ACB$ ...(SSA Congruency theorem)
Hence, now, part(iii) is proved, too.

Now, DC = AB ...(By CPCT)

Multipling both sides by $\frac{1}{2}$ , we get,
($\frac{1}{2}$ ) DC = $\frac{1}{2}$ ) AB

Now, as we know, CM = ($\frac{1}{2}$ ) DC
Therefore, CM = ($\frac{1}{2}$ ) AB
Hence, part (iv) is proved.

## NCERT solutions for class 9 Maths Chapter 7 Triangles Exercise 7.2

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

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In $\mathrm{?}ABC$, we have

AB = AC ...(Given)
Also, $\mathrm{?}B$ = $\mathrm{?}C$ ...(Since corresponding angles of equal sides are equal)

Multipling both sides by $\frac{1}{2}$ , we get,
($\frac{1}{2}$ ) $\mathrm{?}B$ = ($\frac{1}{2}$ ) $\mathrm{?}C$
$?$ $\mathrm{?}OBC$ = $\mathrm{?}OCB$

Also, it is given that, OB and OC are bisectors of $\mathrm{?}B$ and $\mathrm{?}C$, respectively,

Therefore, $\mathrm{?}OBA$ and $\mathrm{?}OCA$
Therefore, OB = OC
...($?$ corresponding sides of equal angles are equal)

Hence, part (i) is proved. In $\mathrm{?}ABO$ and $\mathrm{?}ACD$, we have,

AB = AC ...(Given)
$\mathrm{?}OBA$ = $\mathrm{?}OCA$ ...(from part(i))
OB = OC ...(proved earlier)

Therefore, $\mathrm{?}ABO$ $?$ $\mathrm{?}ACO$ ...(By SAS congruency test)

Thus, $\mathrm{?}BAO$ = $\mathrm{?}CAO$ ...(By CPCT)

Hence, AO bisects $\mathrm{?}A$ is proved.

Q2 ) In the $\mathrm{?}ABC$, AD is the perpendicular bisector of BC (see figure).
Show that $\mathrm{?}ABC$ is an isosceles triangle in which AB = AC. NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In $\mathrm{?}ABC$ and $\mathrm{?}ACD$, we have,

DB = DC ...(given)
$\mathrm{?}ADB$ = $\mathrm{?}ADC$ ...(Since, AD is the perpendicular bisector of BC)

and AD is the Common side.

Therefore, $\mathrm{?}ABD$ $?$ $\mathrm{?}ACD$ ...(By SAS congruency test)

Therefore, AB = AC ...(By CPCT)

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

Q3 ) ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure).
Show that these altitudes are equal. NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In $\mathrm{?}ABE$ and $\mathrm{?}ACF$ , we have,

$\mathrm{?}AEB$ = $\mathrm{?}AFC$ ...(BE and CF are perpendiculars drawn to sides AC and AB respectively)

Also, $\mathrm{?}A$ is common angle.
and AB = AC ...(Given)

Therefore, $\mathrm{?}ABE$ $?$ $\mathrm{?}ACF$ ...(By AAS congruency test)

Therefore, AB = AC ...(By CPCT)

Hence, BE = CF ...(By CPCT)
Hence, proved.

Q4 ) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i) $\mathrm{?}ABE$ $?$ $\mathrm{?}ACF$
(ii) AB = AC i.e., ABC is an isosceles triangle. NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In $\mathrm{?}ABE$ and $\mathrm{?}ACF$ , we have,

$\mathrm{?}AEB$ = $\mathrm{?}AFC$($?BEandCFareperpendiculartosidesACandAB\right){\text{\(}}\mathrm{?}BAE$ = $\mathrm{?}CAF$ ...($?{\text{\(}}\mathrm{?}A$ is the Common angle)

and BE = CF ...(Given)

$?$ $\mathrm{?}ABE$ $?$ $\mathrm{?}ACF$ ...(By AAS Congruency test)

Thus, AB = AC ...(By CPCT)

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

Q5 ) ABC and DBC are isosceles triangles on the same base BC (see figure). Show that $\mathrm{?}ABD$ = $\mathrm{?}ACD$. NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In $\mathrm{?}ABC$, we have,

AB = AC ...($?$ $\mathrm{?}ABC$ is an isosceles triangle)

$?$ $\mathrm{?}ABC$ = $\mathrm{?}ACB$ ...(i)($?$ angles opposite to equal sides are equal)

Similarly, in $\mathrm{?}DBC$, we have,

BD = CD ...($?$ $\mathrm{?}DBC$ too, is an isosceles triangle)

$?$ $\mathrm{?}DBC$ = $\mathrm{?}DCB$ ...(ii) ($?$ angles opposite to equal sides are equal)
Now, On adding, Equations (i) and (ii), we get,

$?\mathrm{?}ABC+\mathrm{?}DBC=\mathrm{?}ACB+\mathrm{?}DCB$
$?$ $\mathrm{?}ABD$ = $\mathrm{?}ACD$
Hence, proved.

Q6 ) $\mathrm{?}ABC$ is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure).
Show that $\mathrm{?}BCD$ is a right angle. NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In $\mathrm{?}ABC$, we have,

AB = AC ...(i)(Given)
$\mathrm{?}ACB$ = $\mathrm{?}ABC$ ...(ii)($?$ angles opposite to equal sides are equal)

From (i) and (iii), AC = AD

Now, in $\mathrm{?}ADC$ , we have,

$\mathrm{?}ACD$ = $\mathrm{?}ADC$ ...($?$ angles opposite to equal sides are equal)

Also, $\mathrm{?}ACD$ = $\mathrm{?}BDC$ ...(iv)($?\mathrm{?}ADC=\mathrm{?}BDC$)

On adding Equations (ii) and (iv), we get,

$?\mathrm{?}ACB+\mathrm{?}ACD=\mathrm{?}ABC+\mathrm{?}BDC$
$?$ $\mathrm{?}BCD$ = $\mathrm{?}ABC$ + $\mathrm{?}BDC$

Adding $\mathrm{?}BCD$ on both sides, we have,

$?\mathrm{?}BCD+\mathrm{?}BCD=\mathrm{?}ABC+\mathrm{?}BDC+\mathrm{?}BCD$
$?$ 2 $\mathrm{?}BCD$ = ${180}^{?}$ ...($?$ sum of all angles of a triangle is ${180}^{?}$)

Therefore, $\mathrm{?}BCD$ = ${90}^{?}$
Hence, proved.

Q7 ) ABC is a right angled triangle in which $\mathrm{?}A$ = ${90}^{?}$ and AB = AC, find $\mathrm{?}B$ and $\mathrm{?}C$. NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In $\mathrm{?}ABC$, we have,

AB = AC ...(Given)
$\mathrm{?}B$ = $\mathrm{?}C$ ...(i)($?$ angles opposite to equal sides are equal)

Now, we know that,

$\mathrm{?}A$ + $\mathrm{?}B$ + $\mathrm{?}C$ = ${180}^{?}$
$?$ ${90}^{?}$ + $\mathrm{?}B$ + $\mathrm{?}C$ = ${180}^{?}$ ...(given)
$?$ ${90}^{?}$ + $\mathrm{?}B$ + $\mathrm{?}B$ = ${180}^{?}$ ...(from(i))
$?$ 2 $\mathrm{?}B$ = ${180}^{?}$ - ${90}^{?}$
$?$ 2 $\mathrm{?}B$ = ${90}^{?}$
$?$ $\mathrm{?}B$ = ${45}^{?}$
$?$ $\mathrm{?}C$ = ${45}^{?}$, too.

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

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Let $\mathrm{?}ABC$ be an equilateral triangle, such that

AB = BC = CA (by property) Now, we have,
AB = AC
$\mathrm{?}B$ = $\mathrm{?}C$ ...(i)($?$ angles opposite to equal sides are equal)

Similarly,
CB = CA

$?$ $\mathrm{?}A$ = $\mathrm{?}B$ ...(ii)($?$ angles opposite to equal sides are equal)

$\mathrm{?}A$ + $\mathrm{?}B$ + $\mathrm{?}C$ = ${180}^{?}$ ...(iii)($?$ the sums of all angles of a triangle are ${180}^{?}$)

From Equations (i),(ii) and (iii), we have,
$\mathrm{?}A$ + $\mathrm{?}A$ + $\mathrm{?}A$ = ${180}^{?}$
$?$ 3 $\mathrm{?}A$ = ${180}^{?}$
$?$ $\mathrm{?}A$ = ${60}^{?}$
$?\mathrm{?}A=\mathrm{?}B=\mathrm{?}C={60}^{?}$
Hence, proved.

## NCERT solutions for class 9 Maths Chapter 7 Triangles Exercise 7.3

Q1 ) $\mathrm{?}ABC$ and $\mathrm{?}DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that
(i) $\mathrm{?}ABD$ $?$ $\mathrm{?}ACD$
(ii)$\mathrm{?}ABP$ $?$ $\mathrm{?}ACP$
(iii) AP bisects $\mathrm{?}A$ as well as $\mathrm{?}D$
(iv) AP is the perpendicular bisector of BC. NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Given: $\mathrm{?}ABC$ and $\mathrm{?}DBC$ are two isosceles triangles having same base BC, such that AB = AC and BD = CD. Proof:
In $\mathrm{?}ABD$ and $\mathrm{?}ACD$, we have,

AB = AC ...(Given)
BD = CD ...(Given)
Therefore, $\mathrm{?}ABD$ $?$ $\mathrm{?}ACD$ ...(By SSS Congruency test)
In $\mathrm{?}ABP$ and $\mathrm{?}ACP$, we have,
Also, $\mathrm{?}a$ = $\mathrm{?}b$ ...($?$ $\mathrm{?}ABD$ $?$ $\mathrm{?}ACD$)
Therefore, $\mathrm{?}ABP$