My Math Forum Ratio of Areas inside a very interesting triangle

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May 14th, 2015, 09:35 AM   #1
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Ratio of Areas inside a very interesting triangle

Does somebody know how to solve this problem? I don't even know how to start...
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Last edited by skipjack; May 14th, 2015 at 06:10 PM.

 May 14th, 2015, 10:50 AM #2 Senior Member     Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 I am strongly recommending to make the picture, it will help you to understand. (1) The ∆ABH and ∆AHM have the same base AH. The height of the ∆ABH BP is twice greater then the height of the ∆ABH that we draw from the vertex M because of AM=MC. So, the ratio of the areas is 2:1. (2) BP:PC=1 because the ∆ABC is isosceles. Thanks from matisolla
 May 14th, 2015, 06:08 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,617 Thanks: 2072 A diagram would help, as you seem to be using an incorrect diagram. I got 4:1 and 2:1. Thanks from matisolla
May 14th, 2015, 07:25 PM   #4
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Quote:
 Originally Posted by matisolla I don't even know how to start...

We have:

$\displaystyle \Delta ABM,\ m(\hat{A})=90^0,\ AH \perp BM\ (H \in BM),\ AB = 2 AM .$

We'll see soon enough the triangles ABH and MAH.

These two triangles are similar, with ratio $\displaystyle \dfrac{2}{1}.$

What can we say about their areas ?

(The answer is simple, if we just draw a geometrical figure)

Last edited by skipjack; May 15th, 2015 at 04:00 AM. Reason: to correct typo

 May 15th, 2015, 04:01 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,617 Thanks: 2072 There's a standard theorem that gives the ratio of the areas. Thanks from aurel5 and matisolla

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