Question

In the figure, PM is the perpendicular bisector of the segment AB. If C is some point on PM, then $△ABC$ is triangle.

• A. equilateral
• B. obtuse
• C. scalene
• D. right-angled

Answer 1

The correct option is D right-angled

In $△ACO$ and $△COB$,
$AO=OB$ [PM divides AB into two equal parts]
$\angle AOC=\angle COB={90}^{\circ }$ [PM is the perpendicular bisector]
$CO=CO$ [Common side]
Hence, $△ACO$ and $△COB$ are congruent.

Therefore, $\angle ACO=\angle OCB$ [CPCT].

As $\angle ACO={45}^{\circ }$, $\angle ACB=\angle ACO+\angle OCB\angle ACB=2\angle ACO\angle ACB=2\times {45}^{\circ }\angle ACB={90}^{\circ }$

Thus, $△ACB$ is a right-angled triangle.