# Prove that the diagonals of a square are equal and perpendicular to each other

$\\ \text{Given :- ABCD is a square.} \\ To proof :- A C=B D and A C \perp B D \\ Proof :- In \triangle \mathrm{ADB} and \triangle \mathrm{BCA} \\ \mathrm{AD}=\mathrm{BC}[ Sides of a square are equal ] \\ \angle \mathrm{BAD}=\angle \mathrm{ABC}\left[90^{\circ} \mathrm{each}\right] \\ \mathrm{AB}=\mathrm{BA}[\mathrm{Common} side ] \\ \triangle \mathrm{ADB} \cong \triangle \mathrm{BCA}[ SAS congruency rule ] \\ \Rightarrow \mathrm{AC}=\mathrm{BD}[ Corresponding parts of congruent triangles are equal ] \\ \ln \triangle \mathrm{AOB} and \triangle \mathrm{AOD} \\ \mathrm{OB}=\mathrm{OD}[ Square is also a parallelogram \\ therefore, diagonal of parallelogram bisect each other ] \\ \mathrm{AB}=\mathrm{AD}[ Sides of a square are equal ] \\ \mathrm{AO}=\mathrm{AO}[\mathrm{Common} side ]$

$\\ \triangle \mathrm{AOB} \cong \triangle \mathrm{AOD}[\mathrm{SSS} congruency rule ] \\ \Rightarrow \angle \mathrm{AOB}=\angle \mathrm{AOD}[ Corresponding parts of congruent triangles are equal] \\ \angle \mathrm{AOB}+\angle \mathrm{AOD}=180^{\circ}[ Linear pair ] \\ \angle \mathrm{AOB}=\angle \mathrm{AOD}=90^{\circ} \\ \Rightarrow A O \perp B D \\ \Rightarrow A C \perp B D$

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