Question

In a triangle ABC, AB=AC, D is a point on AB such that AD=DC=BC. Show that angle BAC = 36°

Answer 1

Refer to the attached image.

In the given triangle ABC,

it is given that CD = BC

As, angles opposite to the equal sides are always equal.

So, $\angle DBC = \angle BDC = x$ (Let)

As, AB = AC

So, $\angle ABC = \angle ACB = x$

In triangle ABC,

By angle sum property of the triangle ABC, we get

$\angle ABC+ \angle BCA+ \angle BAC = 180^\circ$

$\angle BAC+ x+ x= 180^\circ$

$\angle BAC= 180^\circ - 2x$

As, AD=CD

So, $\angle ACD = \angle DAC$

sO, $\angle DAC = 180^\circ-2x$

Now, consider $\angle BCD = \angle ACB - \angle ACD = x-(180-2x) = 3x-180^\circ$

By angle sum property of the triangle BCD, we get

$\angle BDC+ \angle DBC+ \angle BCD = 180^\circ$

$x+ x+ 3x= 180^\circ$

5x = 180

$x = 72^\circ$

Now, $\angle BAC = 180-2x = 180-2(72) = 36^\circ$

Hence, Proved.

Answer 2

Step-by-step explanation:

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