As shown in the given figure two chord AB and CD of the same circle are parallel to each other. P is the centre of the circle
Show that $∠ C P A = ∠ D P B$

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Solution

$In △ PCD, PC = PD i . e ., △ PCD is an isosceles triangle . Hence, ∠ 1 = ∠ 2 [Angles oppsite to equal sides (PC and PD) are equal] ∠ 2 = ∠ 3 (Correcponding angles) ∠ 1 = ∠ 4 (Correcponding angles) ∴ ∠ 3 = ∠ 4 . . . (i) Similarly, in △ APB, we have : AP = BP i . e ., △ APB is an isosceles triangle . Hence, ∠ 7 = ∠ 8 [Angles oppsite to equal sides (AP and BP) are equal] . . . . . (i i) ∠ 5 = 180 ° - ∠ 3 ∠ 6 = 180 ° - ∠ 4 ∴ ∠ 5 = ∠ 6 (from (i)) . . . . (iii)$

$Now, ∠ 5 + ∠ 7 + ∠ 9 = ∠ 6 + ∠ 8 + ∠ 10 (Angle sum property) \Rightarrow ∠ 6 + ∠ 8 + ∠ 9 = ∠ 6 + ∠ 8 + ∠ 10 (from (ii) and (iii)) \Rightarrow ∠ 9 = ∠ 10$

∴ $∠ C P A = ∠ D P B$

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As shown in the given figure two chord AB and CD of the same circle are parallel to each other. P is the centre of the circle Show that ∠CPA=∠DPB

As shown in the given figure two chord AB and CD of the same circle are parallel to each other. P is the centre of the circle
Show that $∠ C P A = ∠ D P B$