# Thread: Proof: Right Triangle and a Perpendicular

1. ## Proof: Right Triangle and a Perpendicular

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Hypothesis: MN Perpendicular to CB
AM = MC
Thesis: BN^2= CN^2 + AB^2

I've thought this could be resolved using some similarity theorem. In fact ABC is similar to MNC because they've got three corresponding pairs of angles. So CM/BC = MN/AB = CN/AC = k. However I don't know what to do now!

Thank you very much!

2. ## Re: Proof: Right Triangle and a Perpendicular

Originally Posted by goby
[/URL]
Hypothesis: MN Perpendicular to CB
AM = MC
Thesis: BN^2= CN^2 + AB^2

I've thought this could be resolved using some similarity theorem. In fact ABC is similar to MNC because they've got three corresponding pairs of angles. So CM/BC = MN/AB = CN/AC = k. However I don't know what to do now!

Thank you very much!
$\displaystyle BC^2=AC^2+AB^2$

$\displaystyle (BN+CN)^2=(2CM)^2+AB^2$

$\displaystyle BN^2+CN^2+2BN*CN=4CM^2+AB^2$ ----------------- (1)

By similar triangles,

$\displaystyle \frac{CM}{BC}=\frac{CN}{AC}$

$\displaystyle \frac{CM}{BN+CN}=\frac{CN}{2CM}$

$\displaystyle 2CM^2=BN*CN+CN^2$

Substituting this in (1),

$\displaystyle BN^2+CN^2+2BN*CN=2BN*CN+2CN^2+AB^2$

$\displaystyle BN^2=CN^2+AB^2$

QED

3. ## Re: Proof: Right Triangle and a Perpendicular

Yeah, it was easier than I thought! Thank you again!
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