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Question
ABCD is quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.
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Solution
Given: In quadrilateral ABCD, AB = AD and CB = CD.
Construction: Join AC and BD.
To prove: AC is the perpendicular bisector of BD.
Proof: In ΔABC and ΔADC,
AB = AD ...[Given]
BC = CD ...[Given]
And AC = AC ...[Common side]
∴ ΔABC ≅ ΔADC ...[By SSS congruence rule]
⇒ ∠1 = ∠2 ...[By CPCT]
Now, in ΔAOB and ΔAOD,
AB = AD ...[Given]
⇒ ∠1 = ∠2 ...[Proved above]
And AO = AO ...[Common side]
∴ ΔAOB ≅ ΔAOD ...[By SAS congruence rule]
⇒ BO = DO ...[Bt CPCT]
And ∠3 = ∠4 [By CPCT] ...(i)
But ∠3 + ∠4 = 180° ...[Linear pair axiom]
∠3 + ∠3 = 180° ...[From equation (i)]
⇒ 2∠3 = 180°
⇒ ∠3 = `(180^circ)/2`
∴ ∠3 = 90°
i.e., AC is perpendicular bisector of BD.
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