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In a figure, DE II QR, AP and BP are bisectors of angle EAB and angle RBA respectively. Find angle APB.

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Still not, can't read it :(



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See diagram enclosed

Let angle EAP = angle BAP = x    (AP bisector)
Let angle RBP = angle PBA = y    BP is bisector

angle EAB + angle ABR = 180 degrees  as AE and BR are parallel

2 * x + 2 * y = 180 deg        x + y = 90

angle APB = 180 - x - y  in trianle APB
angle APB = 180 - 90 = 90 deg

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