The following figure shows a triangle ABC in which P, Q and R are mid-points of sides AB, BC and CA respectively. S is mid-point of PQ. Prove that :

ar.(△ ABC) = 8 × ar.(△ QSB)

In △ ABC,

R and Q are mid-points of AC and BC respectively.

∴ RQ || AB (By mid-point theorem)

We know that,

Area of triangles on the same base and between the same parallel lines are equal.

△ PBQ and △ PAR lie on same base (AP = BP) and between same parallel lines RQ and AB.

∴ Area of △ PBQ = Area of △ APR ……………(1)

P and R are mid-points of AB and AC respectively.

∴ PR || BC (By mid-point theorem)

∴ PR || BQ

∴ PBQR is a parallelogram.

PQ is the diagonal.

⇒ Area of △ PBQ = Area of △ PQR (Diagonal of parallelogram divides it into two triangles of equal areas.) ………(2)

From equation (1) and (2), we get :

⇒ Area of △ PBQ = Area of △ PQR = Area of △ APR ……..(3)

P and Q are mid-points of AB and BC respectively.

∴ PQ || AC (By mid-point theorem)

∴ PQ || RC

∴ PQCR is a parallelogram.

RQ is the diagonal.

⇒ Area of △ PQR = Area of △ RQC (Diagonal of parallelogram divides it into two triangles of equal areas.) ………(4)

From equation (3) and (4), we get :

⇒ Area of △ PBQ = Area of △ PQR = Area of △ APR = Area of △ RQC = x (let).

From figure,

⇒ Area of △ PBQ + Area of △ PQR + Area of △ APR + Area of △ RQC = Area of △ ABC

⇒ x + x + x + x = Area of △ ABC

⇒ 4x = Area of △ ABC

⇒ x = $\dfrac{1}{4}$ Area of △ ABC

⇒ Area of △ PBQ = $\dfrac{1}{4}$ Area of △ ABC ……..(5)

In △ PBQ,

S is the mid-point of PQ and BS is median.

⇒ Area of △ QSB = Area of △ PSB (Median divides triangle into two triangles of equal area)

⇒ Area of △ QSB = $\dfrac{1}{2}$ Area of △ PBQ

⇒ Area of △ PBQ = 2 Area of △ QSB ……….(6)

Substituting value of area of △ PBQ from equation (6) in (5), we get :

⇒ 2 Area of △ QSB = $\dfrac{1}{4}$ Area of △ ABC

⇒ Area of △ QSB = $\dfrac{1}{8}$ Area of △ ABC.

Hence, proved that area of △ QSB = $\dfrac{1}{8}$ area of △ ABC.