Question

In the given construction of a $\Delta ABC,$ what is the value of $\angle CAB?$

Answer 1

From the figure, In $\angle AXB,$ PQ is the perpendicular bisector of AX; the two triangles formed are congruent by SAS rule.
AB = XB
$\Rightarrow \angle BAX=\angle BXA$
$\angle ABC=\angle BAX+\angle BXA\text{(exterior angle property)}$
$\angle ABC=2\angle BXA$
$\angle ABC=\angle LXY\left(\angle AXB\text{is angle bisectorof}\angle LXY\right)$
$\angle ABC={75}^{\circ }$
Similarly,
$\angle ACB=\angle MYX={60}^{\circ }$
In $\Delta ABC,$ we have
$\angle A+\angle B+\angle C={180}^{\circ }$
$\angle A+{75}^{\circ }+{60}^{\circ }={180}^{\circ }$
$\angle A={180}^{\circ }-{135}^{\circ }$
$\angle A={45}^{\circ }$
Thus, $\angle CAB={45}^{\circ }$.