# 08 Circles with diameters equal to corresponding sides of the triangle

**Problem**

From the figure shown below, O_{1}, O_{2}, and O_{3} are centers of circles located at the midpoints of the sides of the triangle ABC. The sides of ABC are diameters of the respective circles. Prove that

where A_{1}, A_{2}, A_{3}, and A_{4} are areas in shaded regions.

**Solution**

a = length of side AB

b = length of side BC and

c = length of side AC

$A_{O1} = \frac{1}{4}\pi a^2$ → Area of circle with diameter AB

$A_{O2} = \frac{1}{4}\pi b^2$ → Area of circle with diameter BC

$A_{O3} = \frac{1}{4}\pi c^2$ → Area of circle with diameter AC

$A_4 = A_{O3} - A_{O1} + A_1 - A_{O2} + A_2 + A_3$

$A_4 = \frac{1}{4}\pi c^2 - \frac{1}{4}\pi a^2 + A_1 - \frac{1}{4}\pi b^2 + A_2 + A_3$

$A_4 = (\frac{1}{4}\pi c^2 - \frac{1}{4}\pi a^2 - \frac{1}{4}\pi b^2) + A_1 + A_2 + A_3$

$A_4 = \frac{1}{4}\pi (c^2 - a^2 - b^2) + (A_1 + A_2 + A_3)$

$A_4 = \frac{1}{4}\pi \, [ \, c^2 - (a^2 + b^2) \, ] + (A_1 + A_2 + A_3)$

${AB}^2 + {BC}^2 = {AC}^2$

$a^2 + b^2 = c^2$

Thus,

$A_4 = \frac{1}{4}\pi \, [ \, c^2 - c^2 \, ] + (A_1 + A_2 + A_3)$

$A_4 = A_1 + A_2 + A_3$ (*okay!*)