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April 3rd, 2011, 08:50 AM   #1
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Another geometry proof

Hi,

I need a help with this one:

We have triangle ABC. D lies on AC and size of AC = CD. S is the point of intersection of AB and ACB angle bisector. M is midpoint of AB. Intersection of DM and CS is K. I need to help with proof that SB = KB (or that angles BSK and SKB are equal).
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January 29th, 2018, 05:59 AM   #2
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This problem is not true for every triangle; take a triangle rectangle in C and the angles A=30º and B=60º. Then the angle BSK is equal to 105º, so the angle BKS will be less than 75º.

Last edited by skipjack; January 29th, 2018 at 02:32 PM.
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January 29th, 2018, 02:53 PM   #3
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That seems to be an invalid argument, euzebio, as K would lie inside triangle ABC for your example.

However, what has to be proved doesn't hold in your example anyway.
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February 4th, 2018, 10:36 AM   #4
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Hi skipjack, I just showed a counter example. Of course it's true for some triangle. This statement is for all triangles, but I show that is not true.

Last edited by skipjack; February 4th, 2018 at 12:33 PM.
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February 4th, 2018, 12:22 PM   #5
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Your counterexample is valid (SB and KB aren't equal), but the reasoning you gave is flawed, as K lies inside the triangle ABC that you described.
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February 11th, 2018, 04:13 PM   #6
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Exercise full proof

Taken from GeoSolver
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February 13th, 2018, 01:17 AM   #7
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Are the angles CAB and CBK equal as the diagram states?

If so, it is quite simple (find the similar triangles)

Last edited by daredman; February 13th, 2018 at 01:19 AM.
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