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...

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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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