Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and $E$, respectively, such that

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Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and $E$, respectively, such that

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$Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and $E$, respectively, such that $\angle EAC = \angle ACD = 90^\circ$. Prove that $EB \cdot BD = AB \cdot BC$$

Mellie Oct 31, 2015

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I think she is saying that ABC has a right angle at B. Legs AB and CB are extended past point B to points D and E, respectively, such that <EAC=<ACD=90*. Prove that EB * BD = AB * BC.

Guest Nov 15, 2015

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Guest · Answer 1 · 2015-11-15T01:20:22

I think she is saying that ABC has a right angle at B. Legs AB and CB are extended past point B to points D and E, respectively, such that <EAC=<ACD=90*. Prove that EB * BD = AB * BC.

Melody · Answer 2 · 2015-10-31T05:48:52

Mellie, I have tried to read theis question but I cannot. It keeps going out of the viewing area.

Can you pleas present it in a more friendly form.

Thanks

(I really would like to try and help you)

Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and $E$, respectively, such that

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Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and $E$, respectively, such that

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