Question

Prove that CF is equal to AD in the given figure in which E is the midpoint of AC and CF is drawn parallel to DB?

Answer 1

If the question is prove that:

ar (parallelogram MHOK) =ar (parallelogram ABCD)

Then the solution is:

ABCD is a parallelogram and BD is its diagonal.

Since, a diagonal in a parallelogram divides it in two triangles of equal area,

So, ar(ΔABD) = ar(ΔBCD)

Since ar(ABCD) = ar(ΔABD) + ar(ΔBCD)

⇒ar(ΔABD) = ar(ΔBCD) = ar(ABCD)

Since, diagonals of a parallelogram bisect each other,

So, AO is the median of ΔABD.

So, ar(ΔAOD) =ar(ΔABD)

Now, M is the midpoint of AD and MH is parallel to OD,

So, by using the converse of mid point theorem, H is the mid point of AO.

In ΔAOD, OM is the median.

Since a median divides the triangle in two triangles of equal area,

ar(ΔAMO) = ar(ΔOMD) =ar(ΔAOD) =ar(ΔABD) = ar(ABCD) =ar(ABCD)

Now, MH is the median for ΔAOM

So, ar(ΔMHO) = ar(ΔMKO) =ar(ΔAOM) =ar(ABCD) =ar(ABCD)

ar (MHOK) =ar(ΔMHO) + ar(ΔMKO) =ar(ABCD) =ar(ABCD)

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