NCERT Solutions for Class 9 Maths Chapter 7 Triangles

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

Answer :

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

AC = AD ....(Given)

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

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

Also, AB is a common side,

So, we can say,
$\mathrm{?}ABC$ ${\displaystyle ?}$ $\mathrm{?}ABD$ ...(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) $\mathrm{?}ABD$ ${\displaystyle ?}$ $\mathrm{?}BAC$
(ii) BD = AC
(iii) $\mathrm{?}ABD$ = $\mathrm{?}BAC$

Answer :

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

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

And AB is a common side.

Therefore, $\mathrm{?}ABD$ ${\displaystyle ?}$ $\mathrm{?}BAC$ ...(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.

Answer :

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, $\mathrm{?}AOD$ ${\displaystyle ?}$ $\mathrm{?}BOC$ ...(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 $\mathrm{?}ABC$ ${\displaystyle ?}$ $\mathrm{?}CDA$.

Answer :

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,
$\mathrm{?}ABC$ ${\displaystyle ?}$ $\mathrm{?}CDA$ ...(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) $\mathrm{?}APB$ ${\displaystyle ?}$ $\mathrm{?}AQB$
(ii) BP = BQ or B is equidistant from the arms of $\mathrm{?}A$.

Answer :

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

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

AB is the common side.
$?$ $\mathrm{?}$ {APB}\) ${\displaystyle ?}$ $\mathrm{?}AQB$ ...(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.

Answer :

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

AB = AD ...(Given)
$\mathrm{?}BAD$ = $\mathrm{?}EAC$ ...(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)

$?$ $\mathrm{?}ABC$ ${\displaystyle ?}$ $\mathrm{?}ADE$ ...(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) $\mathrm{?}DAP$ ${\displaystyle ?}$ $\mathrm{?}EBP$
(ii) AD = BE

Answer :

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))

$?$ $\mathrm{?}DAP$ ${\displaystyle ?}$ $\mathrm{?}EBP$ ...(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) $\mathrm{?}AMC$ ${\displaystyle ?}$ $\mathrm{?}BMD$
(ii) $\mathrm{?}DBC$ is a right angle
(iii) $\mathrm{?}DBC$ ${\displaystyle ?}$ $\mathrm{?}ACB$
(iv)CM = (1/2) AB

Answer :

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

To prove :
(i) $\mathrm{?}AMC$ ${\displaystyle ?}$ $\mathrm{?}BMD$
(ii) $\mathrm{?}DBC$ is a right angle
(iii) $\mathrm{?}DBC$ ${\displaystyle ?}$ $\mathrm{?}ACB$
(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)
$?$ $\mathrm{?}AMC$ ${\displaystyle ?}$ $\mathrm{?}BMD$(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, $\mathrm{?}DBC$ ${\displaystyle ?}$ $\mathrm{?}ACB$ ...(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.

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 $\mathrm{?}A$

Answer :

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, $\mathrm{?}ABO$ ${\displaystyle ?}$ $\mathrm{?}ACO$ ...(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.

Answer :

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, $\mathrm{?}ABD$ ${\displaystyle ?}$ $\mathrm{?}ACD$ ...(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.

Answer :

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, $\mathrm{?}ABE$ ${\displaystyle ?}$ $\mathrm{?}ACF$ ...(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) $\mathrm{?}ABE$ ${\displaystyle ?}$ $\mathrm{?}ACF$
(ii) AB = AC i.e., ABC is an isosceles triangle.

Answer :

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

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

and BE = CF ...(Given)

$?$ $\mathrm{?}ABE$ ${\displaystyle ?}$ $\mathrm{?}ACF$ ...(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 $\mathrm{?}ABD$ = $\mathrm{?}ACD$.

Answer :

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.

Answer :

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

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

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

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

AD = AC ...(proved earlier)
$\mathrm{?}ACD$ = $\mathrm{?}ADC$ ...($?$ 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 $\mathrm{?}C$.

Answer :

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.

Answer :

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.

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) $\mathrm{?}ABD$ ${\displaystyle ?}$ $\mathrm{?}ACD$
(ii)$\mathrm{?}ABP$ ${\displaystyle ?}$ $\mathrm{?}ACP$
(iii) AP bisects $\mathrm{?}A$ as well as $\mathrm{?}D$
(iv) AP is the perpendicular bisector of BC.

Answer :

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)
and AD is Common side.

Therefore, $\mathrm{?}ABD$ ${\displaystyle ?}$ $\mathrm{?}ACD$ ...(By SSS Congruency test)
Hence, part (i) is proved.


In $\mathrm{?}ABP$ and $\mathrm{?}ACP$, we have,

AB = AC ...(Given)
Also, $\mathrm{?}a$ = $\mathrm{?}b$ ...($?$ $\mathrm{?}ABD$ ${\displaystyle ?}$ $\mathrm{?}ACD$)
and AP is the common side.

Therefore, $\mathrm{?}ABP$