Homework Help

In the diagram, triangle ABC is inscribed in the circle and FEG is a tangent to the...

user profile pic

inquire123 | Student | (Level 3) eNoter

Posted August 31, 2013 at 11:57 AM via web

dislike 1 like

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.

Thanks

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)

1 Answer | Add Yours

user profile pic

durbanville | High School Teacher | (Level 1) Educator Emeritus

Posted August 31, 2013 at 2:25 PM (Answer #1)

dislike 1 like

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

` `

Sources:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes