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

(i) ∠POC = $\Big(22\dfrac{1}{2}\Big)^{°}$

(ii) ∠BDC = 2 ∠POC

(iii) ∠BOP = 3 ∠COP

(i) Let ∠POC = x°.

We know that,

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

From figure,

⇒ ∠OCP = ∠OBP = $\dfrac{90°}{2}$ = 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° = $\dfrac{45°}{2}$

⇒ x° = $\Big(22\dfrac{1}{2}\Big)^{°}$

⇒ ∠POC = $\Big(22\dfrac{1}{2}\Big)^{°}$.

Hence, proved that ∠POC = $\Big(22\dfrac{1}{2}\Big)^{°}$.

(ii) From figure,

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

⇒ ∠BDC = 2 × $\Big(22\dfrac{1}{2}\Big)^{°}$

⇒ ∠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.