Q11 of 42 Page 220

In the given figure, ABC is a triangle right angled at B. D and E are points on BC trisect it.

Prove that 8AE2 = 3AC2 + 5AD2.


Given, ABC triangle


Need to prove 8AE2 = 3AC2 + 5AD2


⇒ In Δ ABD, ∠B = 90°


∴ AC2 = AB2 + BC2 ….eq(1)


⇒ Similarly, AE2 = AB2 + BE2 …….eq(2)


⇒ And AD2 = AB2 + BD2 …….eq(3)


⇒ Form eq(1)


⇒ 3AC2 = 3AB2 + 3 BC2 …eq(4)


⇒ From eq(2)


⇒ 5AD2 = 5AB2 + 5BD2 ……eq(5)


Adding equation (4) and (5)


3AC2 + 5AD2 = 8AB2 + 3 BC2 + 5BD2


= 8AB2 + 3 + 5


= 8(AB2 + BE2)


= 8AE2


Hence, proved


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