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In Figure, A E bisects /C A D\ a n d\ /B...

In Figure, `A E` bisects `/_C A D\ a n d\ /_B=/_Cdot` Prove that `A E||B C`

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Solution
In `/_\ABC`, we have
ext `/_CAD=/_B+/_C`
⇒ ext.`/_CAD=2/_C [∵/_B=/_C]`
⇒` 2/_CAE=2/_C ` [∵ ext.'/_CAD` is bisected by `AE`]
⇒`/_CAE=/_C⇒/_CAE=/_ACB=AE∥BC`
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