In the adjoining figure, ABC is a right angled triangle right angled at B. Semicircles are drawn on AB, BC and CA as diameter. Show that the sum of areas of semicircles drawn on AB and BC as diameter is equal to the area of the semicircle drawn on CA as diameter.

Since, ABC is a right angled triangle.

Using pythagoras theorem,

AC² = AB² + BC² ………(1)

Area of semi-circle with diameter AB = $\dfrac{π \times \Big(\dfrac{AB}{2}\Big)^2}{2}$

= $\dfrac{AB^2.π}{8}$ ………(2)

Area of semi-circle with diameter BC = $\dfrac{π \times \Big(\dfrac{BC}{2}\Big)^2}{2}$

= $\dfrac{BC^2.π}{8}$ ……….(3)

Area of semi-circle with diameter AC = $\dfrac{π \times \Big(\dfrac{AC}{2}\Big)^2}{2}$

= $\dfrac{AC^2.π}{8}$ ……….(4)

Adding equations (2) and (3) we get,

Area of semi-circle with diameter AB + Area of semi-circle with diameter BC = $\dfrac{AB^2.π}{8} + \dfrac{BC^2.π}{8}$

= $\dfrac{π}{8}(AB^2 + BC^2)$

From Eq 1,

Area of semi-circle with diameter AB + Area of semi-circle with diameter BC = $\dfrac{π}{8}AC^2$

From Eq 4,

Area of semi-circle with diameter AB + Area of semi-circle with diameter BC = Area of semi-circle with diameter AC

Hence, proved that the sum of areas of semicircles drawn on AB and BC as diameter is equal to the area of the semicircle drawn on CA as diameter.