# Parallelograms

## Exercise 9.3 Part 1

Question 1: In the given figure, E is any point on median AD of a ΔABC. Show that ar(ABE) = ar(ACE).

**Answer:** Median divides a triangle into two triangles of equal area because both the triangles have equal base and equal altitude.

Hence, ar(ABD) = ar(ACD)

Similarly, ar(BED) = ar(DEC)

If we remove ΔBEC, i.e. ΔBED + ΔDEC from this, then

ar(ABE) = ar(ACE) proved.

Question 2: In a triangle ABC, E is the mid-point of median AD. Show that ar(BED) = ¼ ar(ABC).

**Answer:** Let us use the figure in previous question to solve this problem.

ED will be the median for ΔBEC

Hence, ar(BED) = ar(CED)

Moreover, ar(BEC) = ½ ar(ABC)

From these equations, it is clear;

ar(BED) = ¼ ar(ABC)

Question 3: Show that the diagonals of a parallelogram divide it into four triangles of equal area.

**Answer:** ABCD is a parallelogram in which diagonals AC and BD intersect at point O.

To Prove: ar(AOB) = ar(AOC) = ar(BOC) = ar(AOD)

As diagonals of a parallelogram bisect each other so, M and N are the mid-points of sides AD and BC respectively. This means; ar(ABNM) = ar(MNCD) = ½ ar(ABCD)

ar(ABO) = ½ ar(ABNM) because triangle on same base and with same height is half in area compared to parallelogram on same base and height.

Similarly, ar(DOC) = ½ ar(MNCD)

This means; ar(ABO) = ar(DOC) = ¼ ar(ABCD)

Similarly, following can be proved:

ar(AOD) = ar(BON) = ¼ ar(ABCD)

Hence, ar(AOB) = ar(BOC) = ar(DOC) = ar(AOD) = ¼ ar(ABCD)

Question 4: In the given figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ABC) = ar(ABD).

**Answer:** Since AO and BO bisect CD hence they are respective medians of triangles ACD and BCD.

Hence, ar(AOC) = ar(AOD)

Similarly, ar(COB) = ar(DOB)

Hence, ar(AOC) + ar(COB) = ar(AOD) + ar(DOB)

Or, ar(ABC) = ar(ABD) proved