InterviewSolution
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InterviewSolution
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In figure, P is the mid-point of side BC of a parallelogram ABCD such that `angleBAP=angleDAP`. Prove that AD = 2CD. ltBrgt |
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Answer» Given In a parallelogram ABCD, P is mid-point of BC such that `angleBAP=angleDAP`. To prove `" "AD=2CD" "` Proof Since, ABCD is a parallelogram. ltBrgt So, AD`||`BC and AB is transversal, then `" "angleA+angleB=180^(@)" "`[sum of cointerior angles is `180^(@)`] `rArr" "angleB=180^(@)-angleA" "...(i)` In `Delta`ABP, `" "anglePAB+angleB+angleBPA=180^(@)" "`[by angle sum property of triangle] `rArr" "(1)/(2)angleA+180^(@)-angleA+angleBPA=180^(@)" "` [from Eq. (i)] `rArr" "angleBPA-(angleA)/(2)=0` `rArr" "angleBPA=(angleA)/(2)" "...(ii)` `rArr" "angleBPA=angleBAP` `rArr" "AB=BP` [opposite sides of equal angles are equal] On multiplying both sides by 2, we get `" "2AB=2BP` `rArr" "2AB=BC" "` [since P is the mid-point of BC] ltBrgt `rArr" "2CD=AD` `" "`[since, ABCD is a parallelogram, then AB=CD and BC =AD] |
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