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Question
Prove that the diagonals of a square are equal and perpendicular to each other.
Solution
Given:
ABCD is a square with diagonals AC and BD intersecting each other at point O.
To prove: AC = BD and ∠AOB = 90°
Proof:
In ΔDAB and ΔCBA,
AD = BC ...(Sides of a square)
∠DAB = ∠CBA ...(Each 90)
AB = AB ...(Common)
⇒ ΔDAB ≅ ΔCBA ...(SAS Congruence)
⇒ AC = BD ...(C.P.C.T)
In ΔAOB and ΔBOC,
AD = BC ...(Sides of a square)
OB = OB ...(Common)
OA = OC ...(Diagonals of a square bisect each other)
⇒ ΔAOB ≅ ΔBOC ...(SSS Congruence)
⇒ ∠AOB =∠BOC ...(C.P.C.T)
But, ∠AOB + ∠BOC = 180° ...(Linear pair)
⇒ ∠AOB = ∠BOC = `(180°)/(2)` = 90°
∴ AC = BD ⇒ Diagonals are equal
And, ∠AOB = 90° ⇒ Diagonals are perpendicular to each other.
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