In a quadrilateral ABCD, $\angle\text{A}+\angle\text{D}=90^\circ.$ Prove that $AC^2 + BD^2 = AD^2 + BC^2.$
[Hint: Produce AB and DC to meet at E]
[Hint: Produce AB and DC to meet at E]
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Given: A quadrilateral ABCD in which $\angle\text{A}+\angle\text{D}=90^\circ.$
To prove: $AC^2 + BD^2 = AD^2 + BC^2$
Construction: Join AC and BD.
Produce AB and BC to meet at E.
Proof: In $\triangle\text{ADE}$
$\angle\text{BAD}+\angle\text{CDA}=90^\circ$ [Given]
$\angle\text{E}=90^\circ$ [Int. angle of a $\triangle$]
By Pythagoras theorem in $\triangle\text{ADE}$ and $\triangle\text{BCR},$
$AD^2 = AE^2 + DE^2 ......(i)$
$\Rightarrow BC^2 = BE^2 + EC^2 ......(ii)$
Adding (i) and (ii), we get
$AD^2 + BC^2 = AE^2 + EC^2 + DE^2 + BE^2 ......(iii)$
By Pythagoras theorem in $\triangle\text{ECA}$ and $\triangle\text{EBD},$
$AC^2 = AE^2 + CE^2 ......(iv)$
$\Rightarrow BD^2 = BE^2 + DE^2 ......(v)$
$\Rightarrow AD^2 + BC^2 = AE^2 + EC^2 + DE^2 + BE^2 .......(vi)$ [Adding (iv) and (v)]
$\Rightarrow AC^2 + BD^2 = AD^2 + BC^2$ [Using (iii)]
Hence, proved.
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