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Suppose quadrilateral $ABCD$ is inscribed in a circle such that $AB=4, BC=5, CD=6,$ and $DA=7.$ Find the length of $AC.$

I do realize that this is a repost, but the previous asker and answerers did not state a clear method as to how to do this problem. I believe Ptolemy's theorem would be useful, but I am unsure how to apply it. Can someone give me a hint please?

markvs
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questionasker
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1 Answers1

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Use the cosine theorem for triangles ABC and ADC. The angles ABC and ADC add up to 180. Denote the angle ABC by $X$. Then

$$4^2+5^2 - 2*4*5\cos X= 6^2+7^2+2*6*7*\cos X,$$

so $124\cos X = -44$. So $\cos X= -44/124=-11/31$, then $AC^2=4^2+5^2-40*11/31$.

markvs
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