Geometry Questions for CAT XAT & MBA

Angle Bisectors between Parallel Lines – Geometry Concept

Table of Contents

🧠 Concept: Angle Bisector Between Parallel Lines

When two parallel lines are intersected by a transversal, they form equal corresponding and alternate interior angles.

In the figure:

\( l \parallel m \)
\( SQ \) is the angle bisector of \( \angle PQO \), so \( \angle PQS = x \)
\( SO \) is the angle bisector of \( \angle QOR \), so \( \angle ROS = y \)

We know that:

\[ \angle PQR + \angle QRO = 180^\circ \Rightarrow 2x + 2y = 180^\circ \Rightarrow x + y = 90^\circ \]

Now in triangle \( \triangle QOS \):

\[ x + y + \theta = 180^\circ \Rightarrow \theta = 90^\circ \]

Conclusion: The angle between the two bisectors \( SQ \) and \( SO \) is \( \theta = 90^\circ \).

🧪 Practice Question

Q1. In the figure below, lines \( l \parallel m \), and segments \( SQ \) and \( SO \) bisect \( \angle PQR \) and \( \angle QRO \) respectively. What is the value of \( \angle QOS \)?

Options:

(a) \( 60^\circ \)
(b) \( 75^\circ \)
(c) \( 90^\circ \)
(d) \( 120^\circ \)

Show Answer & Explanation

Correct Answer: (c) \( 90^\circ \)

Explanation:

Since \( SQ \) and \( SO \) bisect angles that together form a straight line:

\[ \angle PQR + \angle QRO = 180^\circ \Rightarrow 2x + 2y = 180^\circ \Rightarrow x + y = 90^\circ \]

Then in triangle \( \triangle QOS \):

\[ x + y + \theta = 180^\circ \Rightarrow 90^\circ + \theta = 180^\circ \Rightarrow \theta = 90^\circ \]

Hence, \( \angle QOS = 90^\circ \)

Right Triangle and Pythagoras – Geometry Practice

📘 Geometry Question – Right Triangle and Pythagoras

In the figure below, \( AB \parallel CD \), and line \( AC \) intersects both lines. \( AP \) is drawn such that \( PA \) and \( PC \) are angle bisectors of respective angles