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)

Solution:

AB bisects CD at O signifies that O is the mid-point of CD. AO and BO are medians of triangles ADC and BDC.

Also, the median divides the triangle into two triangles of equal areas. 

Consider ΔACD.

Line-segment CD is bisected by AB at O. Therefore, AO is the median of ΔACD.

∴ Area (ΔACO) = Area (ΔADO) ....(1)

Similarly, Considering ΔBCD , BO is the median.

∴ Area (ΔBCO) = Area (ΔBDO) ...( 2)

Adding equation (1) and equation (2), we obtain

Area (ΔACO) + Area (ΔBCO) = Area (ΔADO) + Area (ΔBDO)

⇒ Area (ΔABC) = Area (ΔABD)

☛ Check: Class 9 Maths NCERT Solutions Chapter 9

Video Solution:

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)
 

Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.3 Question 4

Summary:

If ABC and ABD are two triangles on the same base AB, and line-segment CD is bisected by AB at O, then Area of (ABC) = Area of (ABD).

☛ 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.
• D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that: i) BDEF is a parallelogram. ii) ar (DEF) = 1/4 ar (ABC) iii) ar (BDEF) = 1/2 ar (ABC)