Question

Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC at L and AD produced at E (as shown in the figure below). Prove \[EL=2BL\].

Answer 1

Hint: In the above question, we will consider the \[\Delta BMC\] and \[\Delta EMD\] and make them congruent by ASA criterion of congruence. Also, we will have to use the concept of similarity of triangles between \[\Delta BLC\] and \[\Delta ELA\].

Complete step-by-step answer:
We have been given the figure in the question as below:

In \[\Delta BMC\]and \[\Delta EMD\], we have,

\[\angle BMC=\angle EMD\]

\[\because \]Both angles are vertically opposite angles to each other.

Also, it is given that MC = DM.

\[\angle BCM=\angle EDM\]

\[\because \]Both the angles are alternate interior angle as \[AE||BC\] and DC is the
transversal line.

\[\Delta BMC\cong \Delta EMD\] by ASA congruence criterion of triangles.

Hence, BC = DE according to the corresponding parts of the congruent triangle.

AE = AD + DE = BC + BC = 2 BC….(1)

Now in \[\Delta BLC\] and \[\Delta ELA\], we have,

\[\angle BLC=\angle ALM\]

\[\because \]Both the angles are vertically opposite angles.

\[\angle EAL=\angle BCL\]

\[\because \]Both the angles are alternate interior angles.

\[\Delta BLC\sim \Delta ELA\] by AA similarly.

\[\dfrac{BL}{EL}=\dfrac{BC}{AE}\] by using corresponding parts of a similar triangle.

Now, by using equation (1), we get,

\[\begin{align}

  & \dfrac{BL}{EL}=\dfrac{BC}{2BC} \\

 & \dfrac{BL}{EL}=\dfrac{1}{2} \\

 & EL=2BL \\

\end{align}\]

Hence it is proved that \[EL=2BL\].

Note: Just remember the concept of similarity and congruence of triangles as it will help you a lot in these types of questions.
Also, remember the properties of two parallel lines when cut by a transversal line.
Also, once go through the different criterion of congruence and similarity of a triangle.