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प्रश्न
In a square ABCD, diagonals meet at O. P is a point on BC such that OB = BP.
Show that:
- ∠POC = `[ 22 ( 1°)/( 2 ) ]`
- ∠BDC = 2 ∠POC
- ∠BOP = 3 ∠CPO
उत्तर
(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.
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