SOLUTION: In triangle ABC, point P is on side BC such that PA = 13, PB = 14, PC = 4, and the circumcircles of triangles APB and APC have the same radius. Find the area of triangle ABC.

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Question 1155795: In triangle ABC, point P is on side BC such that PA = 13, PB = 14, PC = 4, and the circumcircles of triangles APB and APC have the same radius. Find the area of triangle ABC.
Answer by MathLover1(20850) About Me  (Show Source):
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In triangle ABC, point P is on side BC such that
+PA+=+13,
PB+=+14,
PC+=+4,
and the circumcircle of triangles APB and APC have the same radius. Find the area of triangle ABC.

triangle-ABC.png

Note that AP is a chord of both circles. Since both circles have the same radius, chord AP must subtend the same angle, i.e. < ABP= < ACP.
Thus, triangle ABC is isosceles with AB=AC.
Let+M be the midpoint of BC. We see that
BC=BP%2B+PC=14%2B4=18
so+BM=CM=BC%2F2=9

Then PM=MC-PC=9-4=5.
Since AM is also an altitude, we can apply Pythagoras theorem to right triangle AMP to get AM:
AM=sqrt%2813%5E2-5%5E2%29=sqrt%28169-25%29=sqrt%28144%29=12
Hence, the area of triangle ABC is:
area=%281%2F2%29AM%2ABC
area=%281%2F2%2912%2A18
area=6%2A18
area=108