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April 6th, 2018, 06:48 AM   #21
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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
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April 6th, 2018, 06:58 PM   #22
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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.
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April 6th, 2018, 07:48 PM   #23
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Quote:
Originally Posted by Ganesh Ujwal View Post
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).
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?
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April 6th, 2018, 09:14 PM   #24
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Please leave this thread and answer my new question.
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April 7th, 2018, 07:57 AM   #25
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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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