Circle Geometry Question: Help! (1 Viewer)

bleh1234 · Aug 30, 2015

This question is from Test Paper 3 of Fitzpatrick's 3 unit textbook.
ABCD is a quadrilateral and each of its sides is a tangent to a circle centre O. Prove that AOB + COD = 180.
Thank you very much!!

InteGrand · Aug 30, 2015

bleh1234 said:
This question is from Test Paper 3 of Fitzpatrick's 3 unit textbook.
ABCD is a quadrilateral and each of its sides is a tangent to a circle centre O. Prove that AOB + COD = 180.
Thank you very much!!

$Draw a diagram of $ABCD$ with a circle inside it centred $O$, and let $P,Q,R,S$ be the points on $AB$, $BC$, $CD$ and $DA$ respectively where the circle touches the quadrilateral.$

$Now, consider triangles $APO$ and $ASO$. Clearly, $\angle APO=\angle ASO$ (both are right angles, as the radius of the circle is perpendicular to its tangents at the point of osculation), $AP=AS$ (tangents to the circle from a common external point are equal), and $AO$ is common. Therefore, $\triangle APO \equiv \triangle ASO$ ($RHS$). Therefore, $\angle AOP = \angle AOS =w$, say (corresponding angles in congruent triangles are equal).$

$Similarly, we can show that $\angle SOD = \angle ROD=x$, say, $\angle DOR = \angle COR = y$, and $\angle COQ = \angle BOQ = z$. Now, from the diagram, we can see that by adding these angles up, we have $2w+2x+2y+2z=360^\circ$ (angles at a point). Dividing through by 2, we have $w+x+y+z=180^\circ$. From the diagram, we can see that $\angle AOB + \angle COD =w+x+y+z$ (as $\angle AOB = w+z$ and $\angle COD = x+y$). Therefore, $\angle AOB + \angle COD = w+x+y+z=180^\circ$. \blacksquare$

Circle Geometry Question: Help! (1 Viewer)

bleh1234

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InteGrand

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