Geometry Circle Question

In the diagram below, A, B, C and D lie on the circle with centre O. If \angle{DBC} = 41^{\circ} and \angle{ACD} = 53^{\circ}, determine with reasoning \angle{BAC} and \angle{AOB}

We know OA=OB=OD – radii of the circle.

Which means, \Delta{AOB} is isosceles and \angle{OAB}=\angle{OBA} – equal angles isosceles triangle.

\angle{AOD}=2\angle{ACD} – angle at the centre twice the angle at the circumference.

\angle{AOB}=106^{\circ}

This means \angle{AOB}=74^{\circ} – angles on a straight line are supplementary

\angle{OAD}=\angle{ODA}=37^{\circ} – equal angles isosceles triangle and the angle sum of a triangle.

\angle{DBA}=\angle{DCA}=53^{\circ} – angle at the circumference subtended by the same arc are congruent.

\angle{CAD}=\angle{CBD}=41^{\circ} – angles at the circumference subtended by the same arc are congruent.

\angle{OAB}=53^{\circ} – equal angle isosceles triangle

Hence \angle{BAC}=12^{\circ}

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Filed under Circle Theorems, Finding an angle, Geometry, Year 11 Specialist Mathematics

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