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maxkor
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Inside the triangle $ABC$ is point $P$, such that $BP > AP$ and $BP > CP$. Prove that $\angle ABC$ is acute.
MHB Proving $\angle ABC$ is Acute: Inside the Triangle $ABC$
- Thread starter maxkor
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Inside triangle ABC, point P is defined such that BP is greater than both AP and CP, leading to the conclusion that angle ABC is acute. The relationships x > y and w > v imply that x + w exceeds y + v, establishing that angle ABC must be less than 90 degrees. The sum of angles CAB and ACB is greater than angle ABC, reinforcing that angle ABC is acute. A misunderstanding arose regarding the problem's requirements, with initial confusion about whether the triangle itself was acute. Ultimately, the proof confirms that angle ABC is indeed acute based on the given conditions.
- #2
mrtwhs
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We are given that $x>y$ and $w>v$ therefore $x+w > y+v = \angle ABC$
Now $\angle CAB > x$ and $\angle ACB >w$ so $\angle CAB + \angle ACB > x+w > y+v = \angle ABC$
But $\angle CAB + \angle ACB = 180 - \angle ABC$ so $180 - \angle ABC > \angle ABC$ and $\angle ABC < 90$
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- #4
maxkor
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Aaahh! I was reading the problem wrong. I thought it was asking to show that the triangle ABC was acute. My bad.maxkor said:
-Dan
For the sake of argument, lets assume this is, in fact, a right triangle with a rectangle inscribed in it.
Here is what looks like a very elegant solution:
Triangles A and B are similar
a/4=2/b
ab=8
But what is happening in that second line? Is there a property of similar triangles I'm forgetting?
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