problem with triangular geometry

**ukorov** · November 5th 2009, 08:26 AM

referring to the attached figure:
In $\triangle ABC$ ,
D is mid-point of BC,
$<br /> \angle ABD = \angle DAC = x<br />$
angle ADB = $45^o$

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

**Opalg** · November 5th 2009, 08:47 AM

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

**ukorov** · November 5th 2009, 09:06 AM

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

**newzad** · November 29th 2009, 02:30 PM

Forgot to write x=30

http://geometri-problemleri.blogspot.com/

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