# D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

Question.

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

Solution:

In right angled $\triangle \mathrm{ACE}$,

$\mathrm{AE}^{2}=\mathrm{CA}^{2}+\mathrm{CE}^{2}$ ...(1)

and in right angled $\triangle B C D$,

$\mathrm{BD}^{2}=\mathrm{BC}^{2}+\mathrm{CD}^{2}$ ...(2)

Adding (1) and (2), we get

$\mathrm{AE}^{2}+\mathrm{BD}^{2}=\left(\mathrm{CA}^{2}+\mathrm{CE}^{2}\right)+\left(\mathrm{BC}^{2}+\mathrm{CD}^{2}\right)$

$=\left(\mathrm{BC}^{2}+\mathrm{CA}^{2}\right)+\left(\mathrm{CD}^{2}+\mathrm{CE}^{2}\right)$

$=\mathrm{BA}^{2}+\mathrm{DE}^{2}$

$\therefore \quad \mathrm{AE}^{2}+\mathrm{BD}^{2}=\mathrm{AB}^{2}+\mathrm{DE}^{2}$