Mathematics
In a triangle ABC, AD is perpendicular to BC. Prove that AB2 + CD2 = AC2 + BD2.
Answer
From figure,
Considering right triangle ABD,
By pythagoras theorem,
AB2 = AD2 + BD2 …….(1)
Considering right triangle ACD,
By pythagoras theorem,
AC2 = AD2 + CD2 …….(2)
Subtracting eqn. 2 from 1 we get,
⇒ AB2 - AC2 = AD2 + BD2 - AD2 - CD2
⇒ AB2 - AC2 = BD2 - CD2
⇒ AB2 + CD2 = BD2 + AC2.
Hence, proved that AB2 + CD2 = AC2 + BD2.
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