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प्रश्न
In the given figure, triangle ABC is a right-angled at B. D is the mid-point of side BC. Prove that AC2 = 4AD2 – 3AB2.

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उत्तर
Given: ∠ABC = 90°, D is mid-point of side BC.
To prove: AC2 = 4AD2 – 3AB2.
Proof: Since D is mid-point of side BC,
∴ BD = DC = `1/2` BC ......(i)
Now, in ΔABC, by Pythagoras theorem,
AB2 + BC2 = AC2
⇒ AB2 + (2BD)2 = AC2 .....[Using (i)]
⇒ AB2 + 4BD2 = AC2 ......(iii)
Similarly, in ΔABD, by Pythagoras theorem,
AB2 + BD2 = AD2
⇒ BD2 = AD2 – AB2
Putting the value of BD2 in equation (ii),
AB2 + 4(AD2 – AB2) = AC2
⇒ AB2 + 4AD2 – 4AB2 = AC2
⇒ 4AD2 – 3AB2 = AC2
or AC2 = 4AD2 – 3AB2
Hence proved.
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