Question

In the below figure, the 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 $\frac{BD}{CD}=\frac{BF}{CE}$

Answer 1

Given:- $CE=AE(\because \text{E is the mid-point of AC})\angle AEF=\angle AFE$

To prove:- $\frac{BD}{CD}=\frac{BF}{CE}$

Construction:- Draw $CG\parallel AB$

Proof:-

$\because \angle AEF=\angle AFE$

$\Rightarrow AE=AF(\because \text{Side opposite to equal angles are equal})$

$\because AE=CE\left(\text{Given}\right)$

$\therefore AE=AF=CE$

In $△AFE$ and $△CGE$,

$\angle AEF=\angle CEG\left(\text{Vertically opposite angle}\right)$

$AE=CE(\because \text{G})$iven

$\angle FAE=\angle GCE\left(\text{Alternate angles}\right)$

By ASA congruent rule,

$△AFE\cong △CGE$

$\Rightarrow CG=AF\left(\text{By C.P.C.T.}\right)$

$\Rightarrow CG=CE(\because CE=EA=FA)$

Now, in $△DBF$ and $△DCG$,

$\angle D=\angle D\left(\text{Common}\right)$

$\angle DBF=\angle DCG\left(\text{Corresponding angles}\right)$

By AA similarity,

$△DBF\sim △DCG$

$\Rightarrow \frac{BD}{CD}=\frac{BF}{CG}\left(\text{Corresponding sides of similar triangles are proportional}\right)$

$\Rightarrow \frac{BD}{CD}=\frac{BF}{CE}(\because CG=CE)$

Hence proved.