# Q13   D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.          Prove that $AE^2 + BD^2 = AB^2 + DE^2.$

S seema garhwal

In $\triangle$ ACE, by Pythagoras theorem,

$AE^2=AC^2+CE^2..................1$

In $\triangle$ BCD, by Pythagoras theorem,

$DB^2=BC^2+CD^2..................2$

From 1 and 2, we get

$AC^2+CE^2+BC^2+CD^2=AE^2+DB^2..................3$

In $\triangle$ CDE, by Pythagoras theorem,

$DE^2=CD^2+CE^2..................4$

In $\triangle$ ABC, by Pythagoras theorem,

$AB^2=AC^2+CB^2..................5$

From 3,4,5 we get

$DE^2+AB^2=AE^2+DB^2$

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