Is there a solution for this isoceles triangle problem?

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The discussion centers on proving that in an isosceles triangle ABC with angles BAC at 20 degrees and BDC at 30 degrees, the sides AD and BC are equal when point D lies on side AB. A proposed method involves using the sine rule for triangles, leading to the conclusion that the ratios of the sides are equal. The calculations demonstrate that sin(10)/sin(150) equals sub(20)/sin(80), confirming that AD equals BC. Participants express interest in the validity of the construction method used for the proof. The problem highlights the intriguing nature of geometric relationships in isosceles triangles.
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Again, this is NOT a homework problem.

Here I am posting a second problem from the math book, The Solvable and the Unsolvable,
which I found in the library.

Consider the isoceles triangle ABC and point D is a point on the side AB.

Please see the diagram below and IGNORE the dots which are used to show the positions of the vertices of the isoceles triangle.

......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.

Is it possible to prove? If it is, how do you do it?

any ideas? There are lots of interesting problems in this book.
 
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hi! i tried this problem by construction. But I don't know if it is good to prove by this method.
 

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AD = [AC*sin(10)/sin(150)]
BC = [AC*sub(20)/sin(80)]

and just using a calculator you can show that

sin(10)/sin(150) = sub(20)/sin(80)

So AD = BC.
 
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