The diagonals AC and BD of a rectangle ABCD intersect each other at P.if angle ABD=50 find anle DPC
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The diagonals AC and BD of rectangle ABCD intersect at P. Also, diagonals of rectangle are equal and they bisect each other. But ∠APB and ∠DPC are vertically opposite angles.
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Given, ABCD is a rectangle
We know that the diagonals of rectangle are same and bisect each other
So, we have
AP = BP
∠PAB = ∠PBA (Equal sides have equal opposite angles)
∠PAB = 50o (Since, given ∠PBA = 50o)
Now, in ∆APB
∠APB + ∠ABP + ∠BAP = 180o
∠APB + 50o + 50o = 180o
∠APB = 180o – 100o
∠APB = 80o
Then,
∠DPB = ∠APB (Vertically opposite angles)
Hence,
∠DPB = 80o
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