Is there a solution for this isoceles triangle problem?

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SUMMARY

The isosceles triangle problem presented involves triangle ABC with sides AB and AC equal, angle BAC measuring 20 degrees, and angle BDC measuring 30 degrees. The goal is to prove that side AD equals side BC. The solution utilizes the Law of Sines, demonstrating that AD can be expressed as AD = AC * (sin(10)/sin(150)), while BC is expressed as BC = AC * (sin(20)/sin(80)). The equality of these two expressions confirms that AD equals BC, thus proving the statement definitively.

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  • Understanding of isosceles triangles and their properties
  • Familiarity with the Law of Sines
  • Basic trigonometric functions and their applications
  • Ability to interpret geometric diagrams
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  • Study the Law of Sines in detail
  • Explore advanced properties of isosceles triangles
  • Practice solving geometric proofs involving angles and sides
  • Learn about trigonometric identities and their applications in geometry
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Mathematics students, educators, and anyone interested in geometric proofs and trigonometry will benefit from this discussion.

davedave
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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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