Question
In a quadrilateral $ABCD$, given that $\angle\text{A}+\angle\text{D}=90^\circ.$ Prove that $AC^2 + BD^2 = AD^2 + BC^2.$
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Answer
Given: A quadrilateral ABCD where $\angle\text{A}+\angle\text{D}=90^\circ.$
To prove: $AC^2 + BD^2 = AD^2 + BC^2$
Construction: Extend AB and CD to intersect at O.
Proof: In $\triangle\text{AOD, }\angle\text{A}+\angle\text{O}+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{O}=90^\circ\ [\angle\text{A}+\angle\text{D}=90^\circ]$
Apply Pythagoras Theorem in $\triangle\text{AOC}$ and $\triangle\text{BOD},$
$AC^2 = AO^2 + OC^2$
$BD^2 = OB^2 + OD^2$
$\therefore AC^2 + BD^2 = (AO^2 + OD^2) + (OC^2 + OB^2)$
$\Rightarrow AC^2 + BD^2 = AD^2 + BC^2$
To prove: $AC^2 + BD^2 = AD^2 + BC^2$
Construction: Extend AB and CD to intersect at O.
Proof: In $\triangle\text{AOD, }\angle\text{A}+\angle\text{O}+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{O}=90^\circ\ [\angle\text{A}+\angle\text{D}=90^\circ]$
Apply Pythagoras Theorem in $\triangle\text{AOC}$ and $\triangle\text{BOD},$
$AC^2 = AO^2 + OC^2$
$BD^2 = OB^2 + OD^2$
$\therefore AC^2 + BD^2 = (AO^2 + OD^2) + (OC^2 + OB^2)$
$\Rightarrow AC^2 + BD^2 = AD^2 + BC^2$
This proves the given relation.
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