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April 1st, 2018, 04:03 PM   #11
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Quote:
Originally Posted by Ganesh Ujwal View Post
In Triangle ABC, the internal bisector of $\angle ABC$ and the external bisector of $\angle ACB$ meet at P.
$\displaystyle \angle BAC = 2 \cdot \angle BPC$
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April 1st, 2018, 07:49 PM   #12
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Quote:
Originally Posted by Denis View Post
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.
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File Type: jpg cat.jpg (12.5 KB, 8 views)
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April 2nd, 2018, 04:59 AM   #13
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NO need for that Helen.
That'll only confuse the OP.
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April 2nd, 2018, 08:41 AM   #14
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Quote:
Originally Posted by helen510 View Post
I am sure you can resolve this ques when you see this pic.
Picture is so small. Please upload bigger one.
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April 3rd, 2018, 09:49 AM   #15
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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 ?
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April 5th, 2018, 07:47 PM   #16
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Quote:
Originally Posted by Denis View Post
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?
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April 5th, 2018, 08:03 PM   #17
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NO. angleABC = angleACB = 70

I'll be plain: you are a joker, else you need classroom help...
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April 6th, 2018, 12:53 AM   #18
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Quote:
Originally Posted by Denis View Post
NO. angleABC = angleACB = 70

I'll be plain: you are a joker, else you need classroom help...
Ok answer is 20
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April 6th, 2018, 02:10 AM   #19
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HOW did you get that?
Show your steps...
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April 6th, 2018, 06:25 AM   #20
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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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