# Thread: a difficult geometry proof

1. ## a difficult geometry proof

Here is the problem.

Consider an isoceles triangle ABC.

Please see the diagram below and IGNORE the dots which are used to show the positions of the points. Point D is a point on side AB.

......................A

.................D

..............B__________C

suppose that side AB=side AC. angle BAC=20 degrees and angle BDC= 30 degrees.

Prove that side AD=side BC.

Please help. thanks

2. Originally Posted by davedave
Here is the problem.

Consider an isoceles triangle ABC.

Please see the diagram below and IGNORE the dots which are used to show the positions of the points. Point D is a point on side AB.

......................A

.................D

..............B__________C

suppose that side AB=side AC. angle BAC=20 degrees and angle BDC= 30 degrees.

Prove that side AD=side BC.

Please help. thanks
using law of sines we have $\frac{\sin (20)}{DC}=\frac{\sin (10)}{AD}$ and $\frac{\sin (80)}{DC}=\frac{\sin (30)}{BC}.$ dividing these two relations gives us: $\frac{AD}{BC}=\frac{\sin (80) \sin (10)}{\sin(20) \sin (30)}=\frac{2 \cos(10) \sin(10)}{\sin (20)}=1.$

3. ## non-trigonometric proof

I guess I forgot to state a requirement for this problem.

We should prove it without using trigonometry.

sorry about my oversight