Angle of this isoscleles triangle.

Triangle ABC is isosceles with base AC. Points P and Q are respectively in CB and AB and such that AC = AP = PQ = QB. What is the degree of angle B?

I have the solution, but I don't understand it.

Represent the magnitude of angle B by m. Then, in order, we obtain angle QPB = m, angle AQP = 2m (??? from now on), angle QAP = 2m, angle QPA = 180 - 4m, angle APC = 3m, angle ACP = 3m. Since angle BCA = angle BAC = 3m, the sum of the angles in ABC is m + 3m + 3m = 7m = 180.

Quote:

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Originally Posted by chengbin

Triangle ABC is isosceles with base AC. Points P and Q are respectively in CB and AB and such that AC = AP = PQ = QB. What is the degree of angle B?

I have the solution, but I don't understand it.

Represent the magnitude of angle B by m. Then, in order, we obtain angle QPB = m, angle AQP = 2m (??? from now on), angle QAP = 2m, angle QPA = 180 - 4m, angle APC = 3m, angle ACP = 3m. Since angle BCA = angle BAC = 3m, the sum of the angles in ABC is m + 3m + 3m = 7m = 180.

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Here is the solution :

Hello chengin,
This is a geometry oddity.Triangle ABC cannot be constructed without knowing the angles and length of base.If given such a diagram and the given equalities the angles can be calculated.

ABC= m
QPC =m isosceles
BQP = 180 - 2m triangle totals 180
AQP =180-(180-2m) = 2m supplements
QPA =180 -4m triangle totals 180
APC=180-m-(180-4m) = 3m supplements
ACP = 3m isosceles
Triangle ABC has three angles m,3m,3m total 7m = 180 m=25.7

Yep, quite an interesting triangle.

If we let BC = k, then:
equal sides AB and BC = k / [2SIN(90/7)]