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.
To Prove: `Delta FBE` is isosceles
`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