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in Geometry by (47.4k points)
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In AD⊥BC, prove that AB2 + CD2 = BD2 + AC2.

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Best answer

From ΔADC, we have

AC2 = AD2 + CD2 … (1)

(Pythagoras theorem)

From ΔADB, we have

AB2 = AD2 + BD2 … (2)

(Pythagoras theorem)

Subtracting (1) from (2) we have,

AB2 – AC2 = BD2 – CD2

AB2 + CD2 = BD2 + AC2

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