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Question
In ΔABC, AD is perpendicular to BC. Prove that: AB2 + CD2 = AC2 + BD2
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Solution
Since triangle ABD and ACD are right triangle right-angled at D,
AB2 = AD2 + BD2 ....(i)
AC2 = AD2 + CD2 ....(ii)
Subtracting (ii) and (i), we get
AB2 - AC2 = BD2 - CD2
⇒ AB2 + CD2 = AC2 + BD2.
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