Question

the bisector of an exterior angle at the vertex of an isosceles triangle is parallel to its base.state and prove its converse

Answer 1

Thank you for asking this question. Here is your answer:

AE is the bisector of the exterior angle ∠DAC of the Δ ABC and AE || BC

Now we will solve it further,

AB || BC   (this is given)

∠1 = ∠2    (given)

∠ B = ∠1    (this is the Corresponding angle)

and ∠C = ∠2    (this is an Alternate angle)

= ∠B = ∠C

= AB = AC

 So it is figured out that Δ ABC is an isosceles triangle.

If there is any confusion please leave a comment below.

Answer 2

Given:- Triangle ACD is isoceles..AB ll DC

To prove:- AC = AD

Construction:- Produce DA to E, such that BA bisects angle CAE

Proof:-

Marks angle ADC, ACD,BAC,BAE as angle 3,angle 4,angle 2, angle 1. respectively

Now, AB ll DC. (Given)

Therefore, angle 2 = angle 4. (alternate interior angles) .....(i)

Also, angle 3 = angle 1. (Corresponding angles) .....(ii)

And, angle 1 = angle. 2. (BA bisects angle CAE; as proved by construction) .....(iii)

Hence, From equations (i),(ii), and (iii) we conclude that :-

Angle 1 = Angle 2 = Angle 3 = Angle 4

Angle 3 = angle 4

Therefore, AC = AD. (Sides opposite to equal angles ar equal) Proved.