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April 1st, 2018, 05:03 PM   #11
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Quote:
 Originally Posted by Ganesh Ujwal In Triangle ABC, the internal bisector of $\angle ABC$ and the external bisector of $\angle ACB$ meet at P.
$\displaystyle \angle BAC = 2 \cdot \angle BPC$

April 1st, 2018, 08:49 PM   #12
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Quote:
 Originally Posted by Denis Messy as hell, but basically correct! Now, IF triangle ABC is isosceles: angleABC = angleACB = ? angleABP = ? angleACP = ?
I am sure you can resolve this ques when you see this pic.
Attached Images
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 April 2nd, 2018, 05:59 AM #13 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,785 Thanks: 970 NO need for that Helen. That'll only confuse the OP.
April 2nd, 2018, 09:41 AM   #14
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Quote:
 Originally Posted by helen510 I am sure you can resolve this ques when you see this pic.
Picture is so small. Please upload bigger one.

 April 3rd, 2018, 10:49 AM #15 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,785 Thanks: 970 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, 08:47 PM   #16
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Quote:
 Originally Posted by Denis 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 ?

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, 09:03 PM #17 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,785 Thanks: 970 NO. angleABC = angleACB = 70 I'll be plain: you are a joker, else you need classroom help...
April 6th, 2018, 01:53 AM   #18
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Quote:
 Originally Posted by Denis NO. angleABC = angleACB = 70 I'll be plain: you are a joker, else you need classroom help...
Ok answer is 20

 April 6th, 2018, 03:10 AM #19 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,785 Thanks: 970 HOW did you get that? Show your steps...
 April 6th, 2018, 07:25 AM #20 Global Moderator   Joined: Dec 2006 Posts: 20,095 Thanks: 1905 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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