# In triangle ABC, sides AB and AC are produced to D and E respectively, so that BD = BC = CE. If BE and CD intersect in F, prove that angle BFD = right angle -1/2 (angle A).

We are given `Delta ABC` . We extend AB to D such that BD=BC, and we also extend AC to E so that CE=BC. Letting the intersection of CD and BE be at F, we are to show that `m/_ BFD = 90-1/2m/_ A` .

Let `m/_ A=A, m/_ ABC=B,...

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We are given `Delta ABC` . We extend AB to D such that BD=BC, and we also extend AC to E so that CE=BC. Letting the intersection of CD and BE be at F, we are to show that `m/_ BFD = 90-1/2m/_ A` .

Let `m/_ A=A, m/_ ABC=B, m/_ ACB=C, m/_ BDF=D, m/_ CEF=E` . Note that, since `Delta DBC` is isosceles by construction, `D=m/_BCF` . Also `Delta BCE` is isosceles, so `E=m/_ EBC` .

If we let `m/_BFD=alpha` then we have the following:

`B+E=alpha + D`
`C+D=alpha +E`

So `B+C+E+D=2alpha + D+E` or `B+C=2alpha` .

Also `A+B+C=180 ==> B+C=180-A` .

Thus `180-A=2alpha ==> alpha=90-1/2A` as required.

** B+E is the measure of `/_ABF` which is an exterior angle to `Delta DBF` and thus equal to the sum of the two remote interior angles which are D and `/_BFD=alpha` .

** C+D is the measure of `/_ACF` and thus is an exterior angle to `Delta ECF` and thus equal to the sum of the remote interior angles. Note that `m/_BFD=m/_CFE` since they vertical angles.` `

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