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In the given figure 'P' is the midpoint of the side BC of a parallelogram ABCD,such that LBAP-LDAP. Prove that AD 2CD |
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Answer» Given :ABCD is a parallelogram. P is the mid point of BC and ∠BAP = ∠DAP To prove :AD = 2 CD Proof : Given, ∠BAP = ∠DAP∴ ∠1 = ∠BAP = 1/2 ∠A ...(1) ABCD is a parallelogram,∴ AD || BC (Opposite sides of the parallelogram are equal)∠A + ∠B = 180° (Sum of adjacent interior angles is 180°)∴ ∠B = 180° – ∠A ...(2) In ΔABP,∠1 + ∠2 + ∠B = 180°(Angle sum property)=> 1/2∠A + ∠2 + 180 - ∠A = 180 [Using equations (1) and (2)]=> ∠A - 1/2 ∠A = 0=> ∠A = 1/2 ∠A ...(3)From (1) and (2), we have∠1 = ∠2In ΔABP,∠1 = ∠2∴ BP = AB (In a triangle, equal angles have equal sides opposite to them)=> 1/2 BC = AB (P is the midpoint on BC)=> BC = 2AB⇒ AD = 2CD (Opposite sides of the parallelogram are equal) |
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