Simple Geom Right?

Geometry Level 3

In Triangle $(ABC)$ , $D$ is on side BC such that, Triangle $(ABD)$ is an isosceles triangle with BD = AB. Given that $\angle{BAC} - \angle{ACB} = 40 deg.$ , Find $\angle{DAC}$ .

The answer is 20.

1 solution

Christian Daang
Jan 24, 2015

Solution:

Let $x = \angle{BAC}$ and $y = \angle{ACB}$ so, $\angle{ABC} = 180-(x+y) deg.$

Since Triangle $ABD$ is an isosceles triangle, then, $\angle{BAD} = \angle{BDA} = \cfrac{x+y}{2}$

So, $\angle{DAC} = x - \cfrac{x+y}{2} = \cfrac{x-y}{2}$

But we know that $x-y = 40$

$\therefore$ , $\angle{DAC} = \cfrac{x-y}{2} = \boxed{20 deg.}$

Simple Geom Right?

The answer is 20.

1 solution

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