In the diagram, BP is parallel to CQ. AB = AC = CD, ∠ABP = 19° and ∠QCD = 68°.
Calculate ∠CDA. Show workings clearly and give reasons for your workings. Ans: ___ °

1. Find the angle between $BP$ and $BC$ using parallel lines.

Since $BP \parallel CQ$ and $BCD$ is a straight line,
$\angle PBC = \angle QCD = 68^\circ$
(corresponding angles).

2. Find the base angle of isosceles triangle $ABC$.

$\angle ABC = \angle PBC - \angle ABP$
$\angle ABC = 68^\circ - 19^\circ = 49^\circ$
Since $AB = AC$, triangle $ABC$ is isosceles, so
$\angle ACB = \angle ABC = 49^\circ.$

3. Use triangle $ACD$ to find $\angle CDA$.

Since $B, C, D$ are collinear,
$\angle ACD = 180^\circ - \angle ACB$
$\angle ACD = 180^\circ - 49^\circ = 131^\circ.$
Since $AC = CD$, triangle $ACD$ is isosceles, so
$\angle CAD = \angle CDA.$
Therefore,
$\angle CDA = \frac{180^\circ - 131^\circ}{2} = 24.5^\circ.$