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Showing posts with label similar triangles II. Show all posts

Showing posts with label similar triangles II. Show all posts

Wednesday, July 4, 2012

Similar Triangles II

$\frac{AD}{BD}$ = $\frac{CD}{AD}$

$\textcolor{red}{AD^2 = BD * CD}$

$\frac{AB}{BC}$ = $\frac{DB}{AB}$

$\textcolor{red}{AB^2 = BD * BC}$

$\frac{AC}{BC}$ = $\frac{CD}{AC}$

$\textcolor{red}{AC^2 = CD * CB}$

Posted by someone opposite of Pierre at 9:39 AM No comments:

Labels: Mathcounts, Mathcounts chapter level, Mathcounts practices, similar triangles II

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