**Question 1:**

**In the given figure, if x = y and AB = CB, then prove that AE = CD.**

**Solution:**

Given: AB = BC

and, x^{o} = y^{o}

To prove: AE = CD

Proof: In \(\Delta\)

Exterior \(\angle\)

\(\Rightarrow\)^{o} = \(\angle\)

Again, in \(\Delta\)

x^{o} = \(\angle\)

Since, x^{o} = y^{o} [Given]

So, \(\angle\)

\(\Rightarrow \)

Thus in \(\Delta\)

\(\angle\)

BC = AB [Given]

and, \)

Thus by Angle-Side-Angle criterion of congruence, we have

\(\Delta\)

The corresponding parts of the congruent triangles are equal.

So, CD = AE [Proved]

**Question 2:**

**ABC is a triangle in which AB = AC. If the bisectors of \(\angle\)B and \(\angle\)C meet AC and AB in D and E respectively, prove that BD = CE.**

**Solution:**

Given: A \(\Delta\)

To prove: BD = CE

Proof: In \(\Delta\)

\(\angle\)

and, \(\angle\)

But \(\angle\)

\(\Rightarrow \)

AB = AC [Given]

\(\angle\)

Thus by Angle-Side-Angle criterion of congruence, we have

\(\Delta\)

The corresponding parts of the congruent triangles are equal.

BD = CE [c.p.c.t]

**Question 3:**

**In the adjoining figure, AD is a median of \(\Delta\)ABC. If BL and CM are drawn perpendiculars on AD and AD produced, prove that BL = CM.**

**Solution:**

Given: A triangle in which D is the midpoint of BC and BL \(\perp\)

To prove: BL = CM

Proof: In \(\Delta\)

\(\angle\)^{o} [Given]

\(\angle\)

BD = DC [Given]

Thus by Angle-Angle-Side criterion of congruence, we have

\(\Delta\)

The corresponding parts of the congruent triangles are equal

So, BL = CM [c.p.c.t]

**Question 4:**

**In \(\Delta\)ABC, D is the mid-point of BC. If DL \(\perp\) AB and DM \(\perp\) AC such that DL = DM, prove that AB = AC.**

**Solution:**

Given: In a \(\Delta\)

To prove: AB = AC

Proof: In right angled triangle \(\Delta\)

\(\angle\)^{o}

Hypt. BD = Hypt. CD [Given]

DL = DM [Given]

Thus, by Right Angle-Hypotenuse_side criterion of congruence, we have

\(\Delta\)

The corresponding parts of the congruent triangles are equal.

Therefore, \(\angle\)

In \(\Delta\)

\(\angle\)

\(\Rightarrow \)

**Question 5:**

**In \(\Delta\)ABC, AB = AC and the bisectors of \(\angle\)B and \(\angle\)C meet at a point O. Prove that BO = CO and the ray AO is the bisector of \(\angle\)A.**

**Solution:**

Given: A \(\Delta\)

To prove: In \(\Delta\)

\(\angle\)

and, \(\angle\)

But, \(\angle\)

So, \(\angle\)

Since base angles are equal, sides are equal

\(\Rightarrow \)

Since OB and OC are the bisectors of angles,

\(\angle\)

\(\angle\)

\(\angle\)

\(\Rightarrow \)

Now, in \(\Delta\)

AB = AC [Given]

\(\angle\)

BO = OC [from(1)]

Thus, by Side-Angle-Side criterion of congruence, we have

\(\Delta\)

The corresponding parts of the congruent triangles are equal

Therefore, \(\angle\)

i.e. , AO bisects \(\angle\)

**Question 6:**

**In the given figure, PQR is an equilateral triangle and QRST is a square. Prove that**

**(i) PT = PS**

**(ii) \(\angle\)PSR = 15 ^{o}**

**Solution:**

Given: PQR is an equilateral triangle and QRST is a square.

To prove: PT = PS

and \(\angle\)^{o}

Proof: Since \(\Delta\)

\(\angle\)^{o} and \(\angle\)^{o}

Since QRTS is a square,

\(\angle\)^{o} and \(\angle\)^{o}

In \(\Delta\)

\(\angle\)

= 60^{o} + 90^{o} = 150^{o}

In \(\Delta\)

\(\angle\)

=60^{o} + 90^{o} = 150^{o} ——- (1)

\(\Rightarrow \)

Thus, in \(\Delta\)

PQ = RS [sides of equilateral triangle PQR]

\(\angle\)

QT = RS [sides of square QRST]

Thus, by Side-Angle-Side criterion of congruence, we have

therefore,Â \(\Delta\)

The corresponding parts of the congruent triangles are equal.

Therefore, PT = PS [c.p.c.t]

Now in \(\Delta\)

PR = RS

\(\Rightarrow \)

But \(\angle\)^{o} [from(1)]

So, by angle sum property in \(\Delta\)

\(\angle\)^{o}

\(\Rightarrow \)^{o} + \(\angle\)^{o}

\(\Rightarrow \)^{o} â€“ 150^{o}

\(\Rightarrow \)^{o}/2 = 15^{o}

**Question 7:**

**In the given figure, ABC is a triangle, right angled at B. If BCDE is a square on side BC and ACFG is a square on AC, prove that AD = BF.**

**Solution:**

Given: ABC is a triangle right angled at B. ACFG is a square and BCDE is a square.

To prove: AD = EF

Proof: Since BCDE is a square,

\(\angle\)^{o} ——- (1)

In \(\Delta\)

\(\angle\)

= \(\angle\)^{o} ——- (2)

In \(\Delta\)

\(\angle\)

Since ACFG is a square,

\(\angle\)^{o}

Thus, we have

\(\angle\)^{o} ——- (3)

From (2) and (3), we have

\(\angle\)

Thus in \(\Delta\)

AC = CF [sides of a square]

\(\angle\)

CD = BC [sides of a square]

Thus, by Side-Angle-Side criterion of congruence, we have

Therefore, \(\Delta\)

The corresponding parts of congruent triangles are equal.

So, AD = BF (c.p.c.t)

**Question 8:**

**Prove that the median from the vertex of an isosceles triangle is the bisector of the vertical angle.**

**Solution:**

Given: ABC is a isosceles triangle in which AB = AC and AD is the median through A.

To prove: \(\angle\)

Proof: In \(\Delta\)

AB = AC [Given]

BD = DC [Given]

AD = AD [Common]

Thus by Side-Side-Side criterion of congruence, we have

\(\Delta\)

The corresponding parts of the congruent triangles are equal.

Therefore, \(\angle\)

**Question 9:**

**In the given figure, ABCD is quadrilateral in which AB || DC and P is the mid-point of BC. On producing, AO and DC meet at Q. Prove that**

**(i) AB = CQ**

**(ii) DQ = DC + AB**

**Solution:**

Given ABCD is a quadrilateral in which AB || DC

To prove:

i) AB = CQ

ii) DQ = DC + AB

Proof: In \(\Delta\)

\(\angle\)

\(\angle\)

BP = PC [Given]

Thus by Angle-Angle-Side criterion of congruence, we have

\(\Delta\)

The corresponding parts of the congruent triangles are equal

Therefore, AB = CQ ——– (1)

Now, DQ = DC + CQ

= DC + AB [from (1)]

**Question 10:**

**In the given figure, OA = OB and OP = OQ. Prove that**

**(i) PX = QX**

**(ii) AX = BX**

**Solution:**

Given: OA = OB and OP = OQ

To prove:

(i) PX = QX

(ii) AX = BX

Proof: In \(\Delta\)

OA = OB [Given]

\(\angle\)

OQ = OP [Given]

Thus by Side-Angle-Side criterion of congruence, we have

\(\Delta\)

The corresponding parts of the congruent triangles are equal.

Therefore, \(\angle\)

Thus, in \(\Delta\)

BQ = OB â€“ OQ

and, PA = OA â€“ OP

But, OP = OQ

and OA =OB [Given]

Therefore, we have, BQ = PA ———- (2)

Now consider triangles \(\Delta\)

\(\angle\)

\(\angle\)

BQ = PA [from (2)]

Thus by Angle-Angle-Side criterion of congruence, we have,

Therefore, \(\Delta\)

PX = QX [c.p.c.t]

AX = BX [c.p.c.t]’