Q9 of 27 Page 289

I have drawn a triangle ABC whose height is AD. If AB>AC, let us prove that, AB2-AC = BD2-CD2

Given: In ΔABC, AB>AC


Height, h = AD



To prove: AB2 - AC2 = BD2 - CD2


Proof:


By applying Pythagoras theorem in ΔACD, we have,


AC2 = AD2 + CD2 ……………… (1)


By applying Pythagoras theorem in ΔABD, we have,


AB2 = AD2 + BD2 ……………… (2)


Subtracting eqn. (1) from (2), we get,


AB2 - AC2 = AD2 + BD2 - (AD2 + CD2)


⇒ AB2 - AC2 = AD2 + BD2 - AD2 - CD2


⇒ AB2 - AC2 = BD2 - CD2


Hence proved.


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