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 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 $\Delta{BPC}$ $\widehat{BCP}+\widehat{BPC}+\widehat{PBC} = 180^\circ$ $\widehat{BPC}=180^\circ-(\widehat{BCP}+\widehat{PBC})$ $=180^\circ-(\widehat{BCA}+\frac{180-\widehat{BCA}}{2}+\frac{\widehat{ABC}}{2})$ $=180^\circ-\frac{2\widehat{BCA}+180^\circ-\widehat{BCA}+\widehat{ABC}}{2}$ $=180^\circ-\frac{180^\circ+\widehat{BCA}+\widehat{ABC}}{2}$ $=180^\circ-\frac{180^\circ+180^\circ-\widehat{BAC}}{2}$ $=180^\circ-\frac{180^\circ+180^\circ-40^\circ}{2}$ $=20^\circ$ Last edited by Ganesh Ujwal; April 6th, 2018 at 07:03 PM.
April 6th, 2018, 07:48 PM   #23
Math Team

Joined: Oct 2011

Posts: 14,597
Thanks: 1039

Quote:
 Originally Posted by Ganesh Ujwal I got the answer, I didn't assume the triangle is isosceles like you.
Hey, how come your English suddenly got better?
My last post says ANY triangle (not isosceles).
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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