# Thread: problem with triangular geometry

1. ## problem with triangular geometry

referring to the attached figure:
In $\triangle ABC$,
D is mid-point of BC,
$
\angle ABD = \angle DAC = x
$

angle ADB = $45^o$

Need desperate help to find the value of x. Thanks.

2. The triangles ABC, DAC are similar (equal angles). Deduce that $AC = \sqrt2DC$. Then apply the sine rule in triangle ADC to find that $\sin x = 1/2$.

3. Originally Posted by Opalg
The triangles ABC, DAC are similar (equal angles). Deduce that $AC = \sqrt2DC$. Then apply the sine rule in triangle ADC to find that $\sin x = 1/2$.
alright that sounds easy, so x = 30 degrees