Question

In a square ABCD, diagonals meet at O. P is a point on BC such that OB = BP.

Show that: 

1. ∠POC = `[ 22 ( 1°)/( 2 ) ]`
2. ∠BDC = 2 ∠POC
3. ∠BOP = 3 ∠CPO

Answer 1

(i) Let ∠POC = x°.

We know that,

Each interior angle equals to 90°. Diagonals of square bisect the interior angles.

From figure,

⇒ ∠OCP = ∠OBP = `(90°)/2`​ = 45°.

We know that,

In a triangle, an exterior angle is equal to sum of two opposite interior angles.

∴ ∠OPB = ∠OCP + ∠POC

⇒ ∠OPB = 45° + x°             ...(1)

In △OBP,

⇒ OB = BP         ...(Given)

⇒ ∠OPB = ∠BOP (Angles opposite to equal sides are equal)          ...(2)

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

⇒ ∠BOP = 45° + x°             ...(3)

We know that,

Diagonals of square are perpendicular to each other.

∴ ∠BOC = 90°

⇒ ∠BOP + ∠POC = 90°

⇒ 45° + x° + x° = 90°

⇒ 2x° = 90° - 45°

⇒ 2x° = 45°

⇒ x° = `(45°)/2`

⇒ x° = `(22 1/2)^°`

⇒ ∠POC = `(22 1/2)^°`

Hence, proved that ∠POC = `(22 1/2)^°`

(ii) From figure,

⇒ ∠BDC = 45°                ...(Diagonals of a square bisect the interior angles)

⇒ ∠BDC = 2 × `(22 1/2)^°`

⇒ ∠BDC = 2 × ∠POC

⇒ ∠BDC = 2 ∠POC

Hence, proved that ∠BDC = 2 ∠POC.

(iii) From equation (3),

⇒ ∠BOP = 45° + x°

⇒ ∠BOP = 45° + 22.5°

⇒ ∠BOP = 67.5°

⇒ ∠BOP = 3 × 22.5°

⇒ ∠BOP = 3 × ∠POC

⇒ ∠BOP = 3 ∠POC

Hence, proved that ∠BOP = 3 ∠COP.