April 3rd, 2011, 07:50 AM  #1 
Newbie Joined: Oct 2010 Posts: 27 Thanks: 0  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). 
January 29th, 2018, 04:59 AM  #2 
Newbie Joined: Feb 2015 From: Brazil Posts: 2 Thanks: 0 
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 01:32 PM. 
January 29th, 2018, 01:53 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,937 Thanks: 2210 
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. 
February 4th, 2018, 09:36 AM  #4 
Newbie Joined: Feb 2015 From: Brazil Posts: 2 Thanks: 0 
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 11:33 AM. 
February 4th, 2018, 11:22 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,937 Thanks: 2210 
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.

February 13th, 2018, 12:17 AM  #7 
Newbie Joined: Feb 2018 From: Israel Posts: 1 Thanks: 0 
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 12:19 AM. 

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