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In the triangle shown, what is the degree measure of $\angle ADB$ enter image description here

I have mainly tried angle-chasing, but it turns out that through plain angle-chasing you cannot find $\angle ADB$ or $\angle EBD$. I explored extending $AE$ and $BD$ to their intersection, but that did not help much either. Can someone get me started on how to solve this problem? I have a feeling dropping a line from $B$ to $DC$ might help, but I cannot figure out which line it would be. EDIT: I realized that dropping a line from B to DC which is parallel to EA might work.

Thanks,

Mike

Mike Smith
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1 Answers1

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In these problems there are two methods to pick from: Synthetic geometry or trigonometry. Although synthetic solutions are more beautiful, they are harder to come up with. To obtain a synthetic solution you need to create isosceles triangles until you end up with an equilateral triangle that can be used to solve the problem or can be used to obtain a congruency.

A more practical theorem is Trigonometric Ceva's Theorem. (The proof of this theorem is done by applying the law of sines to triangles $\bigtriangleup$EOD, $\bigtriangleup$EOA, $\bigtriangleup$AOB, $\bigtriangleup$BOD, and multiplying all the obtained expressions.) To apply the theorem just multiply the ratios of the sines of the angles at each vertex respectively. $$\frac{\sin(x)}{\sin(50)}×\frac{\sin(20)}{\sin(60)}×\frac{\sin(50)}{\sin(30)}×\frac{\sin(40)}{\sin(150 - x)}= 1$$ Rearranging, $$2\sin(x)×\frac{\sin(20)}{\sin(60)}×\frac{\sin(40)}{\sin(150 - x)}= 1$$ From here, the rest is simply trigonometry which I am too tired to finish up right now.

krazy-8
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