AB is the diameter of a circle .P is a point on the semi circle APB. AH and BK are perpendiculars from A and B respectively to the tangent at P. Prove that AH+BK=AB
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AH + BK = AB AB is the diameter of a circle .P is a point on the semi circle APB. AH and BK are perpendiculars from A and B respectively to the tangent at P
Step-by-step explanation:
Extend AB & tangent meeting at Q
now in ΔOPQ & Δ BKQ
OP ⊥ PQ & BK⊥PQ
=> BK/OP = BQ/OQ
Similarly
in ΔOPQ & Δ AHQ
OP ⊥ PQ & AH⊥PQ
=> AH/OP = AQ/OQ
BK/OP + AH/OP = BQ/OQ + AQ/OQ
=> (BK + AH)/OP = (OQ - OB)/OQ + (OA + OQ)/OQ
=> (BK + AH)/OP = (2OQ + OA - OB)/OQ
OA = OB = Radius
=> (BK + AH)/OP =2OQ/OQ
=> (BK + AH)/OP = 2
=> BK + AH =2 * OP
OP = radius => 2 * OP = Diameter = AB
=> AH + BK = AB
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AB is a diameter of a circle .AH and BK are perpendiculars from A and B respectively to the tangent at P .Prove that AH + BK= AB
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