Question

In the square PQRS, the diagonals intersect at O. M is a point on PS such that PM = PO. Show that ∠POM = 3∠MOS.

Answer 1

Let’s consider the following square PQRS.

Let ∠MOS = x

Since, diagonals of a square bisect the angles

∴ ∠OSM = ∠OPM = `90^circ/2` = 45°

Using exterior angle property for ΔOMS,

∠OMP = ∠OMS + ∠MOS

∠OMP = ∠OMS + ∠MOS

∠OMP = 45° + x   ...(i)

In ΔOPM,

OP = MP   ...(Given)

∴ ∠OMP = ∠POM   ...(Angles opposite to equal sides are equal)

⇒ ∠POM = 45° + x   ...(ii) [From (i)]

Diagonals of a square are perpendicular to each other.

∴ ∠POM + ∠MOS = 90°

⇒ 45° + x + x = 90°

⇒ x = 22.5° = `(22  1/2)^circ`

And ∠POM = 45° + x   ...[From (ii)]

⇒ ∠POM = 45° + 22.5° = 3 × 22.5° = 3∠MOS

Hence, proved.