Question

in the given figure ae is the bisector of exterior angle dac such that ae parallel to bc prove that abc is an isosceles triangle​

Answer 1

Given:

AE is the bisector of exterior angle DAC , AE || BC.

To Prove:

∆ABC is isoceles ∆.

Proof:

AE bisects $\angle$DAC , hence we can say that :

$\angle DAE = \angle CAE$

Also, AE || BC , hence $\angle ACB=\angle CAE$

Now, we know that sum of two interior angles of a triangle is equal to the exterior angle.

$\therefore \angle ABC + \angle ACB = \angle DAC$

$=>\angle ABC + \angle ACB = \angle DAE+\angle CAE$

$=>\angle ABC + \angle ACB = \angle CAE+\angle CAE$

$=>\angle ABC + \angle ACB = 2\angle CAE$

$=>\angle ABC + \angle ACB = 2\angle ACB$

$=>\angle ABC = 2\angle ACB-\angle ACB$

$=>\angle ABC = \angle ACB$

$=> AB = AC$

Hence ∆ABC is isoceles ∆.

[Hence proved]