Question

(4. In a square ABCD, diagonals meet at 0. P is
a point on BC, such that OB = BP.
Show that:
(i) ZPOC =
22-
(ii) ZBDC = 2 ZPOC
(iii) ZBOP = 3 ZCOP​

Answer 1

${\mathcal{\red{SOLUTION}}}$

To prove :

$\tt{1)\; \angle POC = \Big(22\dfrac{1}{2}\Big)^{\circ}}$

Given : ABCD is a square and OB = BP

⇒ ∠BOP = ∠BPO   .......(1)   (Angles opposite to equal sides are equal)

Also, $\tt{\angle OPB = \dfrac{90^{\circ}}{2} =45^{\circ}}$    .....(2)    (By Symmetry)

Consider Δ OBP    (by angle sum property)

$\tt{\implies \angle BOP + \angle BPO + \angle OBP = 180^{\circ}}$

$\tt{\implies 2\angle BOP + 45^{\circ} = 180^{\circ}\;\;\;.......[using\;(1)\;and\;(2)]}$

$\tt{\implies 2\angle BOP = 135^{\circ}}$

$\tt{\implies \angle BOP = 67.5^{\circ}\;\;\;\;......(3)}$

Now ∠BOC = 90°   [Diagonals of square bisect each other at 90°]

$\tt{\implies \angle BOP + \angle POC = 90^{\circ}}$

$\tt{\implies \angle POC = 90^{\circ} - 67.5^{\circ} = 22.5^{\circ}\;\;\;\;[using\;(3)]}$

$\tt{i.e. \; \angle POC = \Big(22\dfrac{1}{2}\Big)^{\circ}}$

$\tt{2)\; \angle BDC = 2\angle POC}$

Since ABCD is a square.

AB║CD and BD is a transversal.

$\tt{\angle CBD = \angle BDA = 45^{\circ}\;\;\;\;(Alternate\;interior\;angles)}$

$\tt{Also,\; \angle ADC = 90^{\circ}\;\;\;\; (As\;ABCD\;is\;a\;square)}$

$\tt{\implies \angle BDA + \angle BDC = 90^{\circ}}$

$\tt{\implies 45^{\circ} + \angle BDC = 90^{\circ}}$

$\tt{\implies \angle BDC = 45^{\circ}}$

And ∠POC = 22.5°

So, ∠BDC = 2∠POC = 45°