## Inscribed Triangle

Prove that the total area of the regions 1 and 2 below equals the area of the triangle. Regions 3 and 4 are parts of the semi-circle in which the triangle is inscribed, and $\overline{AB}$ and $\overline{BC}$ are diameters for the other two semi-circles.

Source: Julia Robinson Mathematics Festival

PROOF
Area 1 = area of semi-circle $AEB-$ area 3

$=\frac{\pi}{2}\left (\frac{AB}{2}\right )^2-$ area 3

$=\frac{\pi}{2}\times \frac{AB^2}{4}-$ area 3        (1)

Area 2 = area of semi-circle $BFC-$ area 4

$=\frac{\pi}{2}\left (\frac{BC}{2}\right )^2-$ area 4

$=\frac{\pi}{2}\times \frac{BC^2}{4}-$ area 4        (2)

Adding Eq. (1) and Eq. (2) yields

Area 1 + Area 2 = $\left (\frac{\pi}{2}\times \frac{AB^2}{4}\right )+\left (\frac{\pi}{2}\times \frac{BC^2}{4}\right )-$ area 3 $-$ area 4

$=\frac{\pi}{2}\left (\frac{AB^2+BC^2}{4}\right )-$ area 3 $-$ area 4

Triangle $ABC$ is a right triangle because inscribed angle $B$ intercepts an arc of $180^\circ$.

By the Pythagorean theorem, $AB^2+BC^2=AC^2$. Hence, it follows that

Area 1 + Area 2 = $\frac{\pi}{2}\left (\frac{AC^2}{4}\right )-$ area 3 $-$ area 4

The right hand-side of the above equation is precisely the area of triangle ABC.
DONE