Question

in fig. 2.32, abcd is a cyclic quadrilateral such that ab||dc. if acd = 35° and dbc = 72°, calculate the sizes of the angles of the trapezium.

Explanation:

Step1: Use property of angles in the same segment

Angles in the same segment of a circle are equal. Since $\angle ACD = 35^{\circ}$, then $\angle ABD=35^{\circ}$ (angles in the same segment).

Step2: Calculate $\angle ABC$

$\angle ABC=\angle ABD + \angle DBC$. Given $\angle ABD = 35^{\circ}$ and $\angle DBC=72^{\circ}$, so $\angle ABC=35^{\circ}+72^{\circ}=107^{\circ}$.

Step3: Use property of cyclic - quadrilateral

In a cyclic quadrilateral $ABCD$, $\angle ABC+\angle ADC = 180^{\circ}$ (opposite angles of a cyclic quadrilateral are supplementary). So $\angle ADC=180^{\circ}-\angle ABC=180 - 107^{\circ}=73^{\circ}$.

Step4: Use property of parallel lines

Since $AB\parallel DC$, $\angle BAD+\angle ADC = 180^{\circ}$ (co - interior angles of parallel lines). So $\angle BAD=180^{\circ}-\angle ADC=180 - 73^{\circ}=107^{\circ}$, and $\angle BCD$: We already know that in $\triangle BCD$, $\angle DBC = 72^{\circ}$ and $\angle BDC$ and $\angle BAC$ are angles in the same segment. Also, $\angle BCD=180^{\circ}-(72^{\circ}+35^{\circ}) = 73^{\circ}$ (angle - sum property of a triangle).

Answer:

$\angle A = 107^{\circ}$, $\angle B = 107^{\circ}$, $\angle C = 73^{\circ}$, $\angle D = 73^{\circ}$