Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • In Fig., line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and \angle AEF=\angle AFE.Prove that \frac{BD}{CD}=\frac{BF}{CE}.

In Fig, line segment DF intersects the side AC of a triangle ABC at the point E such that E is the mid-point of CA and \angle AEF=\angle AFE. Prove that \frac{BD}{CD}=\frac{BF}{CE}.

Answers (1)

Given: Line segment DF intersects the side AC of a triangle ABC at the point E such that E is the mid-point of CA and \angle AEF=\angle AFE

To prove:-

\frac{BD}{CD}=\frac{BF}{CE}

Construction:- Take point G on AB such that CG\parallel DF

Proof:- E is the mid-point of CA             (given)

\therefore \; \; \; \; \; \; CE = AE

In \Delta ACG, CG\parallel EF and E is mid-point of CA

Then according to the mid-point theorem

  CE = GF \; \; \; \; \; \; \; ....(1)

In \Delta BCG \; and\; \Delta BDF \; \; CG || DF

According to the basic proportionality theorem.

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

\therefore \frac{BC}{CD}=\frac{BG}{GF}

 \frac{BC}{CD}=\frac{BF-GF}{GF}

 \frac{BC}{CD}=\frac{BF}{GF}=-1

 \frac{BC}{CD}+1=\frac{BF}{GF}

\frac{BC}{CD}+1=\frac{BF}{CE}            (use equation (1))    

\frac{BC+CD}{CD}=\frac{BF}{CE}

\frac{BD}{CD}=\frac{BF}{CE}

Hence proved.   

Posted by

infoexpert23

View full answer

Similar Questions

Latest Question