Question

P, Q and R are mid-points of sides AВ, ВС and CD, respectively, of a rhombus ABCD. Show that PQ is perpendicular to QR.

Answer 1

Given:

ABCD is a rhombus. P, Q, and R are mid-points of sides AB, BC, and CD, respectively.

In \[\triangle \text{ABC}\], P and Q are mid-points of sides AB and BC.

∴ \[\text{PQ} \parallel \text{AC}\] (Mid-point theorem)

In \[\triangle \text{BCD}$, Q and R are mid-points of sides BC and CD.

∴ \[\text{QR} \parallel \text{BD}\] (Mid-point theorem)

In quadrilateral OMQN:

\[\text{QM} \parallel \text{ON}\] (Since $\text{PQ} \parallel \text{AC}$)

\[\text{QN} \parallel \text{OM}\] (Since $\text{QR} \parallel \text{BD}$)

∴ \[\text{OMQN}\] is a parallelogram.

\[\angle \text{MON} = 90^\circ\] (Diagonals of a rhombus intersect at right angles)

\[\angle \text{MQN} = \angle \text{MON} = 90^\circ\] (Opposite angles of a parallelogram are equal)

∴ \[\angle \text{PQR} = 90^\circ\]

∴ \[\text{PQ} \perp \text{QR}\] (Hence Proved)