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December 18th, 2014, 07:30 PM   #1
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Similar Triangles Proof

Givens (Attached File)
Triangle ABC with AB = AC
Altitudes AE and CD

2 problems I'm having trouble with

Prove: AF*BE = CF*AD
Prove: AC*DB = BE*CB

How should I solve this? Should I use transitive property to make several triangles similar?
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December 19th, 2014, 09:40 AM   #2
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$\displaystyle \triangle{ADF}\sim\triangle{CEF}\implies\frac{CE}{ AD}=\frac{CF}{AF}\implies\frac{BE}{AD}=\frac{CF}{A F}\implies AF\cdot BE=CF\cdot AD$


$\displaystyle \triangle{AEC}\sim\triangle{CDB}\implies\frac{AC}{ CE}=\frac{CB}{DB}\implies AC\cdot DB=CE\cdot CB\implies AC\cdot DB=BE\cdot CB$
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