in a parallelogram ABCD,AB=2AD and P is the midpoint of CD prove that angle APB =90 degree Share with your friends Share 0 Manbar Singh answered this We have, ABCD as the given parallelogram.Now, AB = DC and AD = BC Opposite sides of ∥gm are equalNow, AB = 2AD Given⇒AD = 12AB.⇒AD = BC = 12AB as, AD = BC ........1Now, P is the mid point of CD, thenDP = PC = 12DC = 12AB as, AB = DC .......2From 1 and 2, we get AD = DPIn ∆ADP, AD = DP⇒∠APD = ∠DAP Angles opposite to equal sides are equal ........3Since, AB∥DC and AP is a transversal, then∠APD = ∠PAB Alternate interior angles ......4From 3 and 4, we get∠DAP = ∠PAB⇒∠DAP = ∠PAB = 12∠A ......5From 1 and 2, we get BC =PCIn ∆BCP, BC = PC⇒∠CPB = ∠PBC Angles opposite to equal sides are equal ........6Since, AB∥DC and BP is a transversal, then∠CPB = ∠PBA Alternate interior angles ......7From 6 and 7, we get∠PBC = ∠PBA⇒∠PBC = ∠PBA = 12∠B ......8Now, ∠A + ∠B = 180° Adjacent angles of ∥gm are supplementary⇒12∠A + 12∠B = 90° ........9In ∆APB,∠APB + ∠PAB + ∠PBA = 180° Angle sum property⇒∠APB + 12∠A + 12∠B = 180° using 5 and 8⇒∠APB + 90° = 180° using 9⇒∠APB = 90° 3 View Full Answer Allen answered this AB = 2AD DC=2AD DP =AD angle DAP= ang dpa DPA & BAP are equal AP is the angle bisector of A ||| y BP is the angle bisector of B angle A+B=180 Therefore angle pab + pba= 90 Therefore angle apb = 90. ( angle sum pro ) -2
