Question

ABD is a triangle .C is any point on BD such that AB=BC and AC=CD.prove that <BAD=3<ADB

Answer 1

The solution is attached as an image

Answer 2

Given :

ABD is a triangle .C is any point on BD such that AB=BC and AC=CD

To Find :

Prove that $\angle BAD =3 \angle ADB$

Solution:

We are given that AB = BC

Property : Opposite angles of equal sides are equal

So,$\angle BCA= \angle BAC ---1$

We are also given that AC = CD

Property : Opposite angles of equal sides are equal

So, $\angle ADC = \angle CAD ---2$

$\angle BAD = \angle BAC + \angle CAD$

Using 1 and 2

$\angle BAD = \angle BCA + \angle ADC ---3$

Exterior angle property of triangle :  An exterior angle of a triangle is equal to the sum of the two opposite interior angles

Using property

Now$\angle ACB = \angle CAD +\angle CDA$

Substitute value in 3

$\angle BAD = \angle CAD +\angle CDA + \angle ADC$

Using 2

$\angle BAD = \angle CDA+\angle CDA + \angle ADC\\\angle BAD = 3\angle CDA\\\angle BAD =3 \angle ADB$

Hence Proved