Question

in the diagram, o is the centre of a circle which passes through the points a, b, c and d. bce and cdf are straight lines. ba is parallel to cd, adf = 80° and oda = 45°. (i) name three other angles on the diagram which are also 80°. (ii) find (a) reflex bod, (b) abo.

Explanation:

Step1: Use property of parallel lines

Since $BA\parallel CD$ and $\angle ADF = 80^{\circ}$, then $\angle BAD=80^{\circ}$ (alternate - interior angles). Also, angles in the same segment of a circle are equal. So, $\angle BCD = 80^{\circ}$ (opposite angles of a cyclic - quadrilateral $ABCD$ are supplementary, and $\angle BAD$ and $\angle BCD$ are opposite angles, and for a cyclic - quadrilateral, $\angle BAD+\angle BCD = 180^{\circ}$ in the non - reflex sense, and also $\angle BAC$ and $\angle BDC$ subtend the same arc $BC$, so $\angle BAC=\angle BDC$. Since $\angle ADF = 80^{\circ}$, and $\angle BDC$ is part of the angles related to the circle and parallel lines, $\angle BAC = 80^{\circ}$).

Step2: Find reflex $\angle BOD$

The central angle is twice the inscribed angle subtended by the same arc. The non - reflex $\angle BOD = 2\angle BAD$. Since $\angle BAD = 80^{\circ}$, non - reflex $\angle BOD=160^{\circ}$. Then reflex $\angle BOD = 360^{\circ}-160^{\circ}=200^{\circ}$.

Step3: Find $\angle ABO$

$OA = OD$ (radii of the same circle), so $\angle OAD=\angle ODA = 45^{\circ}$. Then $\angle OAB=\angle BAD-\angle OAD=80^{\circ}-45^{\circ}=35^{\circ}$. Also, $OA = OB$ (radii of the same circle), so $\triangle OAB$ is isosceles. Then $\angle ABO=\angle OAB = 35^{\circ}$.

Answer:

(i) $\angle BAD$, $\angle BCD$, $\angle BAC$
(ii) (a) $200^{\circ}$
(b) $35^{\circ}$