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D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that $AE^2 + BD^2 = AB^2 + DE^2$.
Given:
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.
To do:
We have to prove that $AE^2 + BD^2 = AB^2 + DE^2$.
Solution:
In $\triangle \mathrm{ABC}$, $\angle \mathrm{C}=90^{\circ}$
This implies, by Pythagoras theorem,
$\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}$.....(i)
In $\triangle \mathrm{DCE}$,
$\mathrm{DE}^{2}=\mathrm{DC}^{2}+\mathrm{CE}^{2}$........(ii)
Adding equations (i) and (ii), we get,
$\mathrm{AB}^{2}+\mathrm{DE}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}+\mathrm{DC}^{2}+\mathrm{CE}^{2}$
$=\mathrm{AC}^{2}+\mathrm{CE}^{2}+\mathrm{DC}^{2}+\mathrm{BC}^{2}$
$=\mathrm{AE}^{2}+\mathrm{BD}^{2}$ (Since $\mathrm{BD}^{2}=\mathrm{DC}^{2}+\mathrm{BC}^{2}$ and $\mathrm{AE}^{2}=\mathrm{AC}^{2}+\mathrm{CE}^{2}$)
Hence proved.
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