P, Q and R are the mid-points of the sides BC, CA and AB respectively of a triangle ABC. PR and BQ meet at X, CR and PQ meet at Y and BC = 8 cm. Calculate the length of XY.

2 cm

Reason

By mid-point theorem,

The line joining mid-point of any two sides of a triangle is parallel and is equal to half of third side.

Since, P and Q are mid-points of sides BC and AC,

∴ PQ || AB and PQ = $\dfrac{1}{2}$ AB

⇒ PQ || BR and PQ = BR [As, R is mid-point of AB, so BR is equal to half of AB]………………(1)

Since, Q and R are mid-points of sides AC and AB,

⇒ QR || BC and QR = $\dfrac{1}{2}$ BC

⇒ QR || BP and QR = BP [As, P is mid-point of BC, so BP is equal to half of BC]………………(2)

Since, P and R are mid-points of sides BC and AB,

PR || AC and PR = $\dfrac{1}{2}$ AC

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

PQ || BR, QR || BP, PQ = BR, QR = BP.

Since, opposite sides of the quadrilateral PQRB, are equal and parallel.

∴ PQRB is a || gm.

From figure,

⇒ PR = $\dfrac{1}{2}AC$ and PR || AC [By mid-point theorem]

⇒ PR || QC and PR = QC [As, Q is mid-point of AC, so QC is equal to half of AC]………………(3)

Also,

⇒ RQ = $\dfrac{1}{2}BC$ and RQ || BC [By mid-point theorem]

⇒ RQ = PC [P is mid-point of BC] and RQ || PC

We know that,

In parallelogram, diagonals bisects each other.

PQRB is a parallelogram. Diagonal PR and QB meet at X. So, we can also say that X is mid point of PR and QB.

PCQR is a parallelogram. Diagonal PQ and CR meet at Y. So, we can also say that Y is mid point of PQ and CR.

In ΔRPQ, X is mid-point of RP and Y is mid-point of PQ, by mid-point theorem :

∴ XY = $\dfrac{1}{2}QR$

⇒ XY = $\dfrac{1}{2} \times \dfrac{1}{2}BC = \dfrac{1}{4} BC$

= $\dfrac{1}{4} \times 8$ = 2 cm.