In the given figure, $AB$, $BC$, $CD$ and $DA$ are tangents to the circle with centre $O$ forming a quadrilateral $ABCD$.

Show that $\angle AOB + \angle COD = 180°$.

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Question

In the given figure, $AB$, $BC$, $CD$ and $DA$ are tangents to the circle with centre $O$ forming a quadrilateral $ABCD$.

Show that $\angle AOB + \angle COD = 180°$.

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Accepted Answer

Step 1: Quadrilateral ABCD with inscribed circle, radii OP, OQ, OR, OS drawn and angles 1-8 marked
Step 2: Join $OP$, $OQ$, $OR$ and $OS$. $	riangle POB \cong 	riangle QOB \Rightarrow \angle 1 = \angle 2$
Step 3: Similarly $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$
Step 4: $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360°$
Step 5: $\Rightarrow 2(\angle 1 + \angle 8 + \angle 4 + \angle 5) = 360° \Rightarrow \angle AOB + \angle COD = 180°$

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