# a difficult geometry proof

• May 1st 2009, 02:54 PM
davedave
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.

• May 1st 2009, 03:38 PM
NonCommAlg
Quote:

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.

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