Question

In a square ABCD diagonals meet at O. P is a point on BC such that OB = BP. Show that angle POC = 22.5 degree

Answer 1

given: ABCD is a square and OB = BP

in the triangle OBP;

∠BOP = ∠OPB

∠OBP = 90/2 = 45 deg [by the symmetry]

∠BOP = (180-45) / 2 = 135 / 2 = 67.5 deg

∠BOC = 90 deg [diagonals of a square bisect at right angle]

∠POC = 90 - 67.5 = 22.5 deg
again BDC is an isosceles triangle with BCD=90 deg.
and BDC=DBC=45 deg.=2*22.5(POC).

Answer 2

Answer:

Step-by-step explanation:

since BP=OB {given}

angle BOP= angle OPB {angles opposite to equal sides are equal}

also, angle B= 90 ° {angle of a square}

∴ angle OBP= 90/2 = 45°

⇒ angle BOP + angle OPB + angle OBP = 180° {angle sum property}

2∠BOP + 45° = 180°

∠BOP = 135/2

∠BOP= 67.5

and also, ∠BOC= 90° {diagonals of a square bisect each other at right angles}

∴ ∠ POC = ∠ BOC - ∠ BOP

               = 90 - 67.5

               = 22.5 °

Hence proved