Question

in the figure seg AD is perpendicular side BC and B-D-C then prove that AB square - BC square=AC square -CD square

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Answer 1

Answer:

since ABD is a right angle triangle

so by Pythagoras theorem we get -

AB2=AD2+BD2

similarly in triangle ADC we have-

AC2=AD2+CD2

BY equating eq 1& 2 we get

AB2-BD2=AC2-CD2

Answer 2

Answer:

In angle ∆ ACD ~ ∆ ADB,

∆ AD is perpendicular to BC

We have two right angle triangle

∆ADB and ∆ACD.

In ∆ ADB

AC²=AD²+CD²

AD²=AC²–CD². ...(1).

In ∆ ADB

AB²=AD²+BD²

Putting the value of AD² in equation. ...(2).

AB²=(AC²–CD²)+BD²

AB²–BD²=AC²–CD²

Hence, it's proved.

AB²–BD²=AC²–CD².