# In a quadrilateral $\mathrm{ABCD}, \angle \mathrm{A}+\angle \mathrm{D}=90^{\circ}$. Prove that $\mathrm{AC}^{2}+\mathrm{BD}^{2}=\mathrm{AD}^{2}+\mathrm{BC}^{2}$ [Hint: Produce $\mathrm{AB}$ and DC to meet at E]

Given:

In a quadrilateral $\mathrm{ABCD}, \angle \mathrm{A}+\angle \mathrm{D}=90^{\circ}$.

To do:

We have to prove that $\mathrm{AC}^{2}+\mathrm{BD}^{2}=\mathrm{AD}^{2}+\mathrm{BC}^{2}$.

Solution:

Produce $AB$ and $CD$ to meet at $E$.

Join $AC$ and $BD$.

In triangle $AED$,

$\angle A+\angle D =90^{\circ}$

Therefore,

$\angle A+\angle D+\angle E =180^{\circ}$

$\angle E=180^{\circ}-(\angle A+\angle D)$

$=90^{\circ}$

$A D^{2}=A E^{2}+D E^{2}$......(i)

In $\triangle B E C$, by Pythagoras theorem,

$B C^{2}=B E^{2}+E F^{2}$.........(ii)

Adding (i) and (ii), we get,

$A D^{2}+B C^{2}=A E^{2}+D E^{2}+B E^{2}+C E^{2}$..........(iii)

In $\triangle A E C$, by Pythagoras theorem,

$A C^{2}=A E^{2}+C E^{2}$.....(iv)

In $\triangle B E D$, by Pythagoras theorem,

$B D^{2}=B E^{2}+D E^{2}$.......(v)

Adding (iv) and (v), we get,

$A C^{2}+B D^{2}=A E^{2}+C E^{2}+B E^{2}+D E^{2}$.......(vi)

From (iii) and (vi), we get,

$A C^{2}+B D^{2}=A D^{2}+B C^{2}$

Hence proved.

Updated on: 10-Oct-2022

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