Question 1:

In the adjoining figure, PA \(\perp\)

Solution:

Given: PA \(\perp\)

To prove: AO = OB and PO = OQ

Proof: In \(\Delta\)

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

PA = QB [Given]

\(\angle\)

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

\(\Delta\)

\(\Rightarrow\)

Thus, we have O is the midpoint of AB and PQ.

Question 2:

Let the line segments AB and CD intersect at O in such a way that OA = OD and OB = OC. Prove that AC = BD but AC may not be parallel to BD.

Solution:

Given: Line segments AB and CD intersect at O such that OA = OD and OB = OC.

To prove: AC = BD

Proof: In \(\Delta\)

AO = OD [Given]

\(\angle\)

OC = OB [Given]

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

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Thus, we have, AC = BD

In case Â Â \(\angle\)

Question 3:

In the given figure, l||m and M is the mid-point of AB. Prove that M is also the mid-point of any line segment CD having its end points at l and m respectively.

Solution:

Given: Two lines l and m are parallel to each other. M is the midpoint of segment AB. The line segment CD meets AB at M.

To prove: M is the midpoint of CD, that is, CM = MD

Proof:

In \(\Delta\)

\(\angle\)

AM = MB [Given]

\(\angle\)

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

\(\Delta\)

Therfore, by corresponding parts of the congruent triangles are equal, we have, CM = MD

Question 4:

In the given figure, AB = AC and OB = OC. Prove that \(\angle\)

Solution:

Given: AB = AC and O is an interior point of the triangle such that OB = OC

To prove: \(\angle\)

Construction: Join AO

Proof:

In \(\Delta\)

AB = AC [Given]

AO = AO [Common]

OB = OC [Given]

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

\(\Delta\)

\(\Rightarrow\)

Question 5:

In the given figure, ABC is a triangle in which AB = AC and D is a point AB. Through D, a line DE is drawn parallel to BC and meeting AC at E. Prove that AD = AE.

Solution:

Given: A \(\Delta\)

AB = AC

and, DE|| BC

To prove: AD = AE

Proof:

Since DE || BC and AB is atransversal.

So, \(\angle\)

Also DE || BC and AC is a transversal

So, \(\angle\)

But, AB = AC [Given]

So, \(\angle\)

as opposite angles are also equal in case sides are equal

So from (i), (ii) and (iii) we have

\(\angle\)

and in \(\Delta\)

Question 6:

In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of \(\Delta\)

Solution:

Given: AX = AY

To prove: CX = BY

Proof: In \(\Delta\)

AX = AY [Given]

\(\angle\)

AC = AB [Two sides are equal]

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

\(\Delta\)

\(\Rightarrow\)

Question 7:

In the given figure, C is the mid-point of AB. If \(\angle\)

Solution:

Given: C is the mid-point of a line segment AB, and D is point such that,

\(\angle\)

and \(\angle\)

To prove: DC = EC

Proof:

In \(\Delta\)

AC = BC [Given]

\(\angle\)

Also, \(\angle\)

Again in \(\angle\)

ext. \(\angle\)

But, \(\angle\)

\(\Rightarrow\)

\(\Rightarrow\)

Thus in \(\Delta\)

\(\angle\)

AC = BC

and, \(\angle\)

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

\(\Delta\)

The corresponding parts of the congruent triangles are equal.

So, DC = CE [by c.c.p.c.t]

Question 8:

In the given figure, BA \(\perp\)

Solution:

Given: AB \(\perp\)

AB = DE and BF = CD

To prove: AC = EF

Proof:

In \(\Delta\)

BC = BF + FC

and, in \(\Delta\)

FD = FC + CD

But, BF = CD [Given]

So, BC = BF + FC

and, FD = FC + BF

\(\Rightarrow\)

So, in \(\Delta\)

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

BC = FD [Proved above]

AB = DE [Given]

Thus, by Right angle-Hypotenuse-Side criterion of congruence, we have

\(\Delta\)

The corresponding parts of the congruent triangle are equal.

So, AC = EF [c.p.c.t]

AE = \(\angle\)

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

\(\Delta\)

The corresponding parts of the congruent triangles are equal.

So, CD = AE [Proved]’