Question 147277: Ive tried this problem 6 different times, and cannot seem to find the answer.
Given triangle ABC with AB = AC, extend segment AB to a point P so that B is between A and P and BP = BC. In the resulting triangle APC, show that angle ACP is exactly three times the size of angle APC. (By the way, notice that extending segment AB does NOT mean the same thing as extending segment BA.)
thnx!!!
Answer by vleith(2983)   (Show Source):
You can put this solution on YOUR website! Let the angle at BAC = a.
Triangle ABC is isosceles, so its base angles are equal. Call those angles b.
We have a+b+b = 180
By extending B to P, with a length equal to BC, we create a second isosceles triangle BPC. The angle at PBC is the supplement of the angle CBA. That angle is b (from above). So angle PBC = 180-b
Since the second triangle also has base angles equal, we know that angle BPC and BCP are the same. Call that angle c.
Then PBC + 2c = 180.
PBC = 180 - 2c
From above PBC = 180-b, so
180-b = 180 - 2c
b/2 = c
The angle at CPB = c = b/2
The angle PCA = c + b = b/2 + b = 3b/2
Which is 3 times CPB
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