Math, asked by geeta56, 1 year ago

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?

geeta56: yes

Answers

Answered by Anonymous

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 ).

bhavsarriyakgmailcom: but how the lines would be parallel..??

Anonymous: 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!

bhavsarriyakgmailcom: ohk

aryan4668: thanks for your answer

Anonymous: You're welcome. Glad to have helped!

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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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the bisector of the exterior angle CAF of ∆ABC intersects the side BC produced at D .show that BA/AC=BD/DC