# ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF (Fig. 9.15), prove that ar (AER) = ar (AFR)

**Solution:**

Given, ABCD is a square

E and F are the midpoints of BC and CD

R is the midpoint of EF

We have to prove that ar(AER) = ar(AFR)

Draw AN perpendicular to EF

So, AN ⟂ EF

AN is the altitude of the triangle AEF

Area of triangle = 1/2 × base × height

Are of triangle AER = 1/2 × (ER) × (AN)

Since R is the midpoint of EF

ER = FR

So, the area of the triangle AER = 1/2 × (FR) × (AN)

= ar(AFR)

Therefore, ar(AER) = ar(AFR)

**✦ Try This: **D, E and F are respectively the mid-points of the sides BC, CA, and AB of a ABC. Prove that ar (DEF) = 1/4 ar (ABC)

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 9

**NCERT Exemplar Class 9 Maths Exercise 9.3 Problem 5**

## ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF (Fig. 9.15), prove that ar (AER) = ar (AFR)

**Summary:**

ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF (Fig. 9.15), it is proven that ar (AER) = ar (AFR)

**☛ Related Questions:**

- O is any point on the diagonal PR of a parallelogram PQRS (Fig. 9.16). Prove that ar (PSO) = ar (PQO . . . .
- ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig. 9.17). AE intersects CD . . . .
- In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn . . . .

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