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Suppose ABC is a scalene right triangle, and P is the point on hypotenuse \(\overline{AC}\) such that \(\angle{ABP} = 45^{\circ}\). Given that AP = 1 and CP = 2, compute the area of  ABC.

 

I tried using the angle bisector therom but didnt get anywhere with it...

 Nov 13, 2018
 #1
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this problem has no possible solution; carry on.

 Nov 13, 2018
 #2
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This math expert has no possible brain. Don't carry on. 

Guest Nov 13, 2018
 #3
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Why didn't angle bisector theorem work?

Since BP bisects the right angle at B, we can designate BC =x and AB =2x. By Pythagoras' theorem we have:

3^2 =x^2 + 2x^2

9 = 5x^2

x =sqrt(9/5)

The area of ABC triangle = 1/2 * x * 2x =1/2 * sqrt(9/5) * 2sqrt(9/5) =9/5 units^2.

 Nov 13, 2018

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