## Not understanding proof

1. The problem statement, all variables and given/known data
THIS PART I UNDERSTAND ( I am just adding it so the pic makes sence).

Let angle AOP traced out be 30 deg. Produce PM to P' making MP' equal to PM.
The two triangles OMP and OMP' have their sides OM and MP' equal to OM and MP and also the contained angles equal.
Therefore OP' = OP, and angle OP'P = angle OPP' = 60 deg.
That the triangle P'OP is equilateral

THIS IS WHERE I DONT UNDERSTAND:
Hence: OP^2 = PP'^2 = 4PM^2 = 4OP^2 - 4a^2
Where OM equals a.
3OP^2 = 4a^2

I understand why OP^2 = PP'^2 but how do they equal 4PM^2 = 4OP^2 - 4a^2 ?
And where did 3OP^2 = 4a^2 come from?

2. Relevant equations

3. The attempt at a solution
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Homework Help
 Quote by Miike012 1. The problem statement, all variables and given/known data THIS PART I UNDERSTAND ( I am just adding it so the pic makes sence). Let angle AOP traced out be 30 deg. Produce PM to P' making MP' equal to PM. The two triangles OMP and OMP' have their sides OM and MP' equal to OM and MP and also the contained angles equal. Therefore OP' = OP, and angle OP'P = angle OPP' = 60 deg. That the triangle P'OP is equilateral THIS IS WHERE I DONT UNDERSTAND: Hence: OP^2 = PP'^2 = 4PM^2 = 4OP^2 - 4a^2 Where OM equals a. 3OP^2 = 4a^2 I understand why OP^2 = PP'^2 but how do they equal 4PM^2 = 4OP^2 - 4a^2 ? And where did 3OP^2 = 4a^2 come from? 2. Relevant equations 3. The attempt at a solution
Note that, M is the midpoint of PP'.

So we have: $$PP' = 2 PM \Rightarrow PP' ^ 2 = 4 PM ^ 2$$.

And, $$4PM ^ 2 = 4 (OP ^ 2 - OM ^ 2)$$, this is just Pythagorean Identity (note that POM is a right triangle).

Hope that you can get it. :)
 Thank you.

## Not understanding proof

Actually no that doesnt make sence to me....
4PM^2 = 4(OP^2 - OM^2)
.
.

First off.. why are you multiplying the right side by 4?
 Mentor He's not. He is just rewriting 4(PM)2 in a different form, using the Theorem of Pythagoras. BTW, "sence" is not a word in English. since - means because, or due to. sense - a means of determining something scents - smells or aromas cents - fractional parts of a dollar.
 Ok... Theorem of Pythagoras is. PM^2 + OM^2 = OP^2 (right?) We noted that PP' = 2PM Thus. 4PM^2 + OM^2 = OP^2 Then.. 4PM^2 = OP^2 - OM^2 Where did the factor of four from the right side come from?

Mentor
 Quote by Miike012 Ok... Theorem of Pythagoras is. PM^2 + OM^2 = OP^2 (right?)
So PM^2 = OP^2 - OM^2.

From the work that VietDao29 showed, PP' = 4PM^2 = ?
 Quote by Miike012 We noted that PP' = 2PM Thus. 4PM^2 + OM^2 = OP^2 Then.. 4PM^2 = OP^2 - OM^2 Where did the factor of four from the right side come from?
 Still didnt help... Ill just ask my teacher. thanks.
 Mentor PM is half of PP', so PP' = 2PM, hence (PP')2 = (2PM)2 = 4(PM)2. That's all it is.