In the diagram, triangle ABC is inscribed in the circle and FEG is a tangent to the circle at E such that EF=EC.FBC is a straight line and EA intersects BC at D.

Show that triangle FBE is isosceles.

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To Prove: `Delta FBE` is isosceles

FE=EC (given)

`therefore /_ BFE = /_ BCE` (isosceles triangle FCE) Note that F is an extension of CB

In circle BACE angle BAE=angle BCE(arc BE subtends equal angles)

`therefore /_BFE = /_BAE` ( BAE = BCE which = BFE)

In `Delta BAE` `/_BAE = /_FEB` ( tan chord theorem. Angle FEB is exterior)

We know that `/_BAE = /_BFE` (proven above)

`therefore /_BFE = /_FEB`

**Ans: Therefore triangle FBE is isosceles**

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