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(2) The bisector of the exterior/ CAF of aAABC, intersects the side BCproduced at D. Show thatBA BAC DC |
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Answer» Given: ABC is a triangle; AD is the exterior bisector of 4A and meets BCproduced at D; BA is produced to F.To prove: AB/AC = BD/DCConstruction: Draw CE||DA to meet AB at E.Proof: In A ABC. CE||AD cut by AC. ∠CAD = ∠ACE (Alternate angles)Similarly CE || AD cut by AB ∠FAD = ∠AEC (corresponding angles)AC = AE (by isosceles △ theorem) Now in △BAD, CE || DAAE=DC(BPT)AB BD But AC = AE (proved above)AC=DCAB=BD orAB/AC = BD/DC (proved). |
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