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Orthocentric triangle length proof - help needed!

The triangle is setup as shown in the picture where the perpendicular heights of AC and BC meet at H (like an orthocentric system). I am given the additional information that angle <ACB is 45 degrees. I am asked to show CH=AB.. How do I start?

Apologies. I initially stated 'centroid' rather thank 'orthocentric' in my title.

Re: Centroid triangle length proof - help needed!

Let D be the unlabeled intersection point on AC and E be the unlabeled point on BC. You are given three angles: ACB, CEA, and BDC. Add up the three angles of a triangle, and you get 180 degrees. Start finding other angles. What do you know about a triangle with two equal angles? What type of triangle is that? That will help you find similar triangles.

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Re: Centroid triangle length proof - help needed!

Attachment 29707

Okay so I've managed to find these similar triangles (it's not to scale). The angles highlighted in green are 45 degrees, other coloured angles are unknown. I'm still a bit unsure how to move from this to CH=AB though.

Re: Centroid triangle length proof - help needed!

I've also noted that BE=EH and AD=DH due to isosceles triangle and similarly, BH+DH=DC and AH+HE=CE

Re: Centroid triangle length proof - help needed!

Compare triangles DCH and ADB. You have the following:

angle DCH = 90 minus the blue angle

angle DBA = 90 minus the blue angle

angle CHD = 180 - 90 - angle(DCH)

angle BAD = 180 - 90 - angle(DBA)

So, put all that together, and you find triangles DCH and ABD are similar triangles. Then as you stated, DH = AD. So, by the properties of similar triangles, $\displaystyle \dfrac{\text{DH}}{\text{AD}} = \dfrac{\text{CH}}{\text{AB}} = 1$

Re: Centroid triangle length proof - help needed!

Okay, thanks, I see it now :) quite simple, not sure why I couldn't see it before! Thanks very much

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Re: Centroid triangle length proof - help needed!

Hi,

I was a tad worried that the above proof did not take into account that H may well be outside the triangle. Dynamic geometry to the rescue. The two cases of H inside and H outside the triangle do produce different figures, but the proof stays the same for either case (different reasons for some steps). Here are the two situations:

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Re: Centroid triangle length proof - help needed!

Hi johng, I don't see how H could not be in the system, it's the point at which the perpendicular heights of the sides meet.. Why would they meet outside?

Maybe I'm missing something?

Re: Centroid triangle length proof - help needed!

Quote:

Originally Posted by

**ejhowells** Hi johng, I don't see how H could not be in the system, it's the point at which the perpendicular heights of the sides meet.. Why would they meet outside?

Maybe I'm missing something?

Look at the second picture johng gave. Look at where the point B is. That is an example of a triangle where the perpendicular heights of the sides meet outside the triangle (this happens when angle ABC is greater than 90 degrees).