Question

In the given construction,

• PQ is the perpendicular bisector AX
• RS is the perpendicular bisector AY
• XA is the angle bisector of the $\angle LXY$
• YA is the angle bisector of the $\angle MYX$

Then find the value of $\angle CAB$.

Answer 1

From construction
AB = XB (Since B is a point on the perpendicular bisector of AX)
$\Rightarrow \angle BAX=\angle AXB$ (angles opposite to equal sides are equal)

$\angle ABC=\angle BAX+\angle AXB=2\angle AXB=\angle LXY$

( $\because$ XA is the bisector of the $\angle LXY$ )

Therefore, $\angle ABC={75}^{\circ }$

Similarly, $\angle ACB=\angle MYX$
Therefore, $\angle BCA={60}^{\circ }$

In $△$ ABC, we have,
$\angle ABC+\angle BCA+\angle CAB={180}^{\circ }$
${75}^{\circ }+{60}^{\circ }+\angle CAB={180}^{\circ }$
Then $\angle CAB={180}^{\circ }-{135}^{\circ }$
Therefore $\angle CAB={45}^{\circ }$