In a triangle A B C , a point D is chosen on B C such that B D : D C = 2 : 5 . Let P be a point on the circumcircle A B C such that ∠ P D B = ∠ B A C . Then P D : P C is :-

Question

In a triangle A B C , a point D is chosen on B C such that B D : D C = 2 : 5 . Let P be a point on the circumcircle A B C such that ∠ P D B = ∠ B A C . Then P D : P C is :-, 2 : 5, 2 : 5, 2 : 7, 2 : 7

MARKS App · Accepted Answer

B D D C = 2 5 ∠ P D B = ∠ B A C = θ let ∠ P C D = α ⇒ ∠ D P C = θ - α ∠ B A C = ∠ B P C = θ (angle in the same segment) ⇒ B P D = θ - ( θ - α ) = α so Δ P C B ~ Δ P D B P C D P = B C P B = P B B D P C D P 2 = B C P B × P B B D = B C B D P C D P = B C B D = 7 λ 2 λ = 7 2 D P P C = 2 7

Search any question & find its solution

Looking for more such questions to practice?