Question

In a square ABCD, diagonals meet at O. P is a point on BC such that OB = BP.
$\angle BOP=3\angle COP$

State whether the above statement is true or false.

• A. True
• B. False

Answer 1

Answer: (A)

The correct option is A True
Given, $ABCD$ is a $Square$ and $OB=BP$

In $△OBP$,

$\angle BOP=\angle OPB$

$\angle OBP=\frac{90}{2}={45}^{\circ }$ --------[by symmetry or diagonals bisect the angles at the vertices, in a square]

We know that, Sum of interior angles of a triangle is equal to ${180}^{\circ }$

$\angle BOP=\frac{180-45}{2}=\frac{135}{2}={67.5}^{\circ }$ ----------(1)

$\angle BOC={90}^{\circ }$ --------[diagonals bisect at right angles, in a square]

$\angle POC=90-67.5={22.5}^{\circ }$ -----------(11)

From (1) and (11)

$\angle BOP=3\angle POC$

Hence its True