Difference between revisions of "Ptolemy's Theorem"

Revision as of 14:35, 14 January 2016

Examining the similar triangles in (c), we have ${\frac {j}{d}}={\frac {b}{p}}$ , so $jp=bd$

Examining the similar triangles in (d), we have ${\frac {j}{c}}={\frac {a}{q}}$ , so $jq=ac$

So $j(p+q)=bd+ac$ ... Q.E.D.

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 [[File:Ptolemy's_Theorem.png]]
-Examining the similar triangles in (c), we have <math>/frac{j}{d} = /frac{b}{p}</math>, so <math>jp = bd</math>
+Examining the similar triangles in (c), we have <math>\frac{j}{d} = \frac{b}{p}</math>, so <math>jp = bd</math>
-Examining the similar triangles in (d), we have <math>/frac{j}{c} = /frac{a}{q}</math>, so <math>jq = ac</math>
+Examining the similar triangles in (d), we have <math>\frac{j}{c} = \frac{a}{q}</math>, so <math>jq = ac</math>
-So <math>j(p+q) = bd + ac</math>
+So <math>j(p+q) = bd + ac</math> ... Q.E.D.