Question

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Answer 1

$\huge\underline\mathfrak{Question :-}$

P and Q are any two points lying in the same segment of a circle, as shown in the adjoining figure . Prove that ∠PBA +∠PAB = ∠QAB + ∠QBA .

To prove :-∠PBA +∠PAB = ∠QAB + ∠QBA

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SOLUTION :-

∠AQB =∠ APB [ •.•According to the theorem of circle that is angle of the same segment of a circle are equal and so through the figure it is clearly seen that both the angles lie on the same base i.e AB ]

∠Q + ∠QAB + ∠QBA = 180° ______(i)

∠P + ∠PAB + ∠PBA = 180° ______(ii)

After equating both the equations (i) and (ii) we get,

∠Q + ∠QAB + ∠QBA = ∠P + ∠PAB + ∠PBA [ Since both are equal to the sum 180° and so both the equations are also equal since their sum is equal ]

=>∠Q - ∠P + ∠QAB +∠QBA =∠PAB+ ∠PBA

=> ∠QAB + ∠QBA = ∠PAB + ∠PBA [•.• ∠Q and ∠P are same so both gets cut ]

∠QAB + ∠QBA = ∠PAB + ∠PBA ( proved )

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