Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

Solution:

Let's draw the given quadrilateral ABCD.

It is given that,

Area (ΔAOD) = Area (ΔBOC)

Now, adding Area (ΔAOB) on both the sides

area (ΔAOB) + area (ΔAOD) = area (ΔAOB) + area (ΔBOC) 

Hence ar (ΔADB) = ar (ΔACB)

According to Theorem 9.3: Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

Therefore, ΔADB and ΔACB, are lying between the parallels lines. i.e., AB || CD as they lie on the same base AB and have equal areas.

Therefore, ABCD is a trapezium.

☛ Check: Class 9 Maths NCERT Solutions Chapter 9

Video Solution:

Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.3 Question 15

Summary:

If diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC), then ABCD is a trapezium.

☛ Related Questions:

• In Fig.9.23, E is any point on median AD of a ∆ ABC. Show that ar (ABE) = ar (ACE).
• In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4 ar(ABC).
• Show that the diagonals of a parallelogram divide it into four triangles of equal area.
• In Fig. 9.24, 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).