April 1st, 2018, 04:03 PM  #11 
Senior Member Joined: Feb 2010 Posts: 683 Thanks: 129  
April 1st, 2018, 07:49 PM  #12  
Newbie Joined: Mar 2018 From: australia Posts: 15 Thanks: 0  Quote:
 
April 2nd, 2018, 04:59 AM  #13 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 883 
NO need for that Helen. That'll only confuse the OP. 
April 2nd, 2018, 08:41 AM  #14 
Senior Member Joined: Aug 2014 From: India Posts: 295 Thanks: 1  
April 3rd, 2018, 09:49 AM  #15 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 883 
Forget it, Ganesh: that will not help you... Once more: CAN YOU answer: IF your triangle ABC is isosceles: angleABC = angleACB = ? angleABP = ? angleACP = ? YES or NO ? 
April 5th, 2018, 07:47 PM  #16  
Senior Member Joined: Aug 2014 From: India Posts: 295 Thanks: 1  Quote:
angleABC = angleACB = 40 (since you are saying triangle is isosceles) My question why you assume triangle is isosceles instead of obtuse or equilateral etc?  
April 5th, 2018, 08:03 PM  #17 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 883 
NO. angleABC = angleACB = 70 I'll be plain: you are a joker, else you need classroom help... 
April 6th, 2018, 12:53 AM  #18 
Senior Member Joined: Aug 2014 From: India Posts: 295 Thanks: 1  
April 6th, 2018, 02:10 AM  #19 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 883 
HOW did you get that? Show your steps... 
April 6th, 2018, 06:25 AM  #20 
Global Moderator Joined: Dec 2006 Posts: 19,287 Thanks: 1681  Bisectors.PNG $\small\angle$ACD = $\small\angle$ABC + 40$^\circ$ Dividing by 2, $\small\angle$PCD = $\small\angle$PBC + 20$^\circ$ Hence $\small\angle$BPC = 20$^\circ$. 

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