April 6th, 2018, 06:48 AM  #21 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 
Well, Skip generously gave you the steps. Now, for any triangle ABC where angleBAC is given: k = angleBAC, v = angleBPC : show that k = 2v Let u = angleABP v = 180  (180  k  u)  (k + 2u)/2 Multiply by 2: 2v = 360  2(180  k  u)  (k + 2u) simplify: 2v = k 
April 6th, 2018, 06:58 PM  #22 
Senior Member Joined: Aug 2014 From: India Posts: 476 Thanks: 1 
I got the answer, I didn't assume the triangle is isosceles like you. In $\Delta{BPC}$ $\widehat{BCP}+\widehat{BPC}+\widehat{PBC} = 180^\circ$ $\widehat{BPC}=180^\circ(\widehat{BCP}+\widehat{PBC})$ $=180^\circ(\widehat{BCA}+\frac{180\widehat{BCA}}{2}+\frac{\widehat{ABC}}{2})$ $=180^\circ\frac{2\widehat{BCA}+180^\circ\widehat{BCA}+\widehat{ABC}}{2}$ $=180^\circ\frac{180^\circ+\widehat{BCA}+\widehat{ABC}}{2}$ $=180^\circ\frac{180^\circ+180^\circ\widehat{BAC}}{2}$ $=180^\circ\frac{180^\circ+180^\circ40^\circ}{2}$ $=20^\circ$ Last edited by Ganesh Ujwal; April 6th, 2018 at 07:03 PM. 
April 6th, 2018, 07:48 PM  #23  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  Quote:
My last post says ANY triangle (not isosceles). The isosceles portion was to help YOU, in order for YOU to get familiar with the process... Are we now finally finished with this charade?  
April 6th, 2018, 09:14 PM  #24 
Senior Member Joined: Aug 2014 From: India Posts: 476 Thanks: 1 
Please leave this thread and answer my new question.

April 7th, 2018, 07:57 AM  #25 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 
Similar to a British Gentleman to a Lady, I bow at the waist and tip my hat to you, and depart never to be seen again... Whoops, my elbow accidentally just hit the "ignore button"... 

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