Question

(a) Find out the unknown $\angle BOD$ in the given quadrilateral ABDC in which $AB\parallel CD$. Given that AD & BC are angle bisectors of $\angle BAC$ & $\angle DCA$ respectively.

(b) What is the value of (x+y) in the figure below?

[4 MARKS]

Answer 1

(a) Steps: 2 Marks
Result: 1 Mark
(b) Solution: 1 Mark

(a) Since $AB\parallel CD$

$\Rightarrow \angle BAC+\angle DCA={180}^{\circ }$

[Sum of Co-interior angles is equal to ${180}^{\circ }$]

$\Rightarrow \frac{\angle BAC}{2}+\frac{\angle DCA}{2}=\frac{{180}^{\circ }}{2}$

$\Rightarrow \angle OAC+\angle OCA={90}^{\circ }$ [AD & BC are angle bisectors]

In $\Delta AOC,$

$\angle AOC+\angle OAC+\angle OCA={180}^{\circ }$

[sum of all the angles in triangle is ${180}^{\circ }$]

$\Rightarrow \angle AOC={180}^{\circ }-{90}^{\circ }={90}^{\circ }$

$\angle AOC=\angle BOD$ [Vertically Opposite Angles]

$\Rightarrow \angle BOD={90}^{\circ }$

(b) The value of an exterior angle of a triangle is equal to the sum of the opposite interior angles.

As the exterior angle value is given to be 120${}^{\circ }$, the sum of the two interior angles is also 120${}^{\circ }$.
Thus, (x+y) = 120${}^{\circ }$