the bisector of the exterior angle CAF of ∆ABC intersects the side BC produced at D .show that BA/AC=BD/DC
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akanshasehgal5:
Are you sure the relation is BD/DC?
yes
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Step-by-step explanation:
Draw the line through B parallel to AC. Let this line meet AD (extended beyond A) in the point E.
BE parallel to CA => ∠BEA = ∠CAD
=> ΔBED is similar to ΔCAD ( equal angles above and common angle at D)
Also,
∠BEA = ∠CAD = ∠DAF ( AD is angle bisector )
= ∠EAB (vertically opposite angles),
so ΔBEA is isosceles => BA = BE.
Therefore
BA / AC
= BE / AC ( BA = BE )
= BD / DC ( similar triangles BED and CAD ).
but how the lines would be parallel..??
BE is parallel to CA because E is on that line that we drew in the first statement "daw the line through B parallel to AC". So that line in parallel to CA because we *chose* the line that is parallel to AC!
ohk
thanks for your answer
You're welcome. Glad to have helped!
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