Question

$△PSR$ is a triangle right angled at S. D is the mid-point of SR. If the bisector of $\angle PSR$ and perpendicular bisector of $SR$ meet at $O$, then triangle $△OSD$ is:

• A. isosceles
• B. equilateral
• C. isosceles right angled
• D. acute-angled

Answer 1

Answer: (C)

isosceles right angled

Since, $OS$ bisects $\angle PSR$, therefore $\angle RSO=\angle OSP={45}^0$.
Now, in $△OSD$ $\angle ODS={90}^0$ and $\angle DSO={45}^0$.
By Angle Sum Property,
$\angle ODS+\angle DSO+\angle DOS={180}^0$

$\Rightarrow {90}^0+{45}^0+\angle DOS={180}^0$

$\Rightarrow \angle DOS={45}^0$.
Now, $\angle DSO=\angle DOS={45}^0$

$\Rightarrow OD=OS$ ( in a triangle, sides opposite to equal angles are equal).
Also, $\angle ODS={90}^0$

Therefore, $△OSD$ is an isosceles right-angled triangle.