+0

# Help on geometry!!

0
138
3
+36

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
+128
0

this problem has no possible solution; carry on.

Nov 13, 2018
#2
0

This math expert has no possible brain. Don't carry on.

Guest Nov 13, 2018
#3
0

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