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  • 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 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}.

Answers (1)

Given: 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

To prove :-

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

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

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

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

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

Then according to mid-point theorem

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

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

According to 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.   

 

 

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infoexpert23

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