Understanding Triangle Proofs: Solving for Unknown Sides

[Miike012]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

 

[VietDao29]

Homework Statement

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?

Homework Equations

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. :)

 

[Miike012]

Thank you.

 

[Miike012]

Actually no that doesn't make sense to me...
4PM^2 = 4(OP^2 - OM^2)
.
.

First off.. why are you multiplying the right side by 4?

 

[Mark44]

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.

 

[Miike012]

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?

 

[Mark44]

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 = ?

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?

 

[Miike012]

Still didnt help... Ill just ask my teacher. thanks.

 

[Mark44]

PM is half of PP', so PP' = 2PM, hence (PP')² = (2PM)² = 4(PM)². That's all it is.