Mathematics
In quadrilateral ABCD, AB = AD and CB = CD. Prove that AC is perpendicular bisector of BD.
Triangles
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Answer
In △ ABC and △ ADC,
⇒ AB = AD (Given)
⇒ BC = CD (Given)
⇒ AC = AC (Common side)
∴ △ ABC ≅ △ ADC (By S.S.S. axiom)
We know that,
Corresponding parts of congruent triangle are equal.
⇒ ∠BAC = ∠DAC
⇒ ∠BAO = ∠DAO
In △ AOB and △ AOD,
⇒ AB = AD (Given)
⇒ AO = AO (Common side)
⇒ ∠BAO = ∠DAO (Proved above)
∴ △ AOB ≅ △ AOD (By S.A.S. axiom)
We know that,
Corresponding parts of congruent triangle are equal.
⇒ ∠BOA = ∠DOA …….(1)
⇒ OB = OD
From figure,
⇒ ∠BOA + ∠DOA = 180° (Linear pair)
⇒ ∠BOA + ∠BOA = 180° [From equation (1)]
⇒ 2∠BOA = 180°
⇒ ∠BOA = = 90°.
∴ AC is perpendicular bisector of BD.
Hence, proved that AC is perpendicular bisector of BD.
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Related Questions
Assertion (A): ΔABD ≅ ΔACE
Reason (R): ∠ADE + ∠ADB = ∠AEC + ∠AED
But AD = AE
⇒ ∠ADE = ∠AED
∴ ∠ADB = ∠AEC
⇒ ∠ABD ≅ ∠AEC
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(b) In △ ABC and △ DEF, ∠B = ∠E = 90°; AC = DF and BC = EF.
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