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Question
In right-angled triangle ABC in which ∠C = 90°, if D is the mid-point of BC, prove that AB2 = 4AD2 − 3AC2.
Solution
We have,
∠C = 90° and D is the mid-point of BC
In ΔACB, by Pythagoras theorem
AB2 = AC2 + BC2
⇒ AB2 = AC2 + (2CD)2 [D is the mid-point of BC]
AB2 = AC2 + 4CD2
⇒ AB2 = AC2 + 4(AD2 − AC2) [In ΔACD, by Pythagoras theorem]
⇒ AB2 = AC2 + 4AD2 − 4AC2
⇒ AB2 = 4AD2 − 3AC2
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