In triangle ABC, point X is on side BC such that AX=13,BX=10,CX=10, and the circumcircles of triangles ABX and ACX have the same radius. Find the area of triangle ABC.
+396 This question was posted a few days ago, but in that version CX = 4.
I'll sketch out a solution for that one. (This new version is marginally easier.)
For a triangle ABC, the sine rule says that
\(\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B} =\frac{c}{\sin C}= 2R,\)
where R is the radius of the circumscribing circle (of the triangle).
Applying that to the triangles ACX and ABX, we have,
\(\displaystyle \frac{AX}{\sin C}=2R_{1} \quad \text{ and } \quad\frac{AX}{\sin B}=2R_{2}\)
and since the two radii are equal, it follows that the angles B and C are equal meaning that the triangle ABC is isosceles.
Drop a perpendicular from A to BC, meeting BC at Y, say, then XY = 3, and calculate AY using Pythagoras.
The area of ABC will then be equal to (1/2)*BC*AY.
