Relationship of circles and polygons

[Harry_the_cat]

So it's proven, yeah?

 

[Ihatdogs10]

So it's proven, yeah?

by similar triangle?

 

[Harry_the_cat]

Well it involved a few things along the way:
1. circle theorem ( radius and tangent meet at right angles)
2. angle rule ( vertically opposite angles are equal)
3. This lead to proving two triangles similar (using AAA here)
4. Once the triangles were shown to be similar, you needed the relationship between sides
5. Finally rearranging an equation involving fractions to one where no fractions appeared.

 

[Ihatdogs10]

Well it involved a few things along the way:
1. circle theorem ( radius and tangent meet at right angles)
2. angle rule ( vertically opposite angles are equal)
3. This lead to proving two triangles similar (using AAA here)
4. Once the triangles were shown to be similar, you needed the relationship between sides
5. Finally rearranging an equation involving fractions to one where no fractions appeared.

Is this all the reason?

 

[Harry_the_cat]

Can you think of anything else?

 

[Ihatdogs10]

I can't

Can you think of anything else?

 

[Harry_the_cat]

I think that's about all.
Good on you for sticking it out till the end. I think you've learnt (and revised) some things along the way!
So you are in Year 10, I gather? Good luck with your senior studies next year.

 

[Ihatdogs10]

so I am going to write it all? in the reason

 

[Ihatdogs10]

I think that's about all.
Good on you for sticking it out till the end. I think you've learnt (and revised) some things along the way!
So you are in Year 10, I gather? Good luck with your senior studies next year.

Yes I am, thank you very much

 

[Harry_the_cat]

You'll need to show your working (including reasons) for each step along the way.

 

[Ihatdogs10]

You'll need to show your working (including reasons) for each step along the way.

okay thank you

 

[TMarvel]

If you know triangular similarities then this is a piece of cake for you!

First join PX and YQ
observe that,
∠YOQ = ∠XOP (vertically opp ∠s) and ∠YQO = ∠XPO = 90°
⟹ΔXPO ~ ΔYQO (by AA similarity)
⟹ PX/QY = PO/QO
⟹PX ⋅ QO = QY ⋅ PO
∴ Proved

Hope it helps!

 

[Harry_the_cat]

Yes TMarvel, that's correct. But the OP needed to UNDERSTAND each step.
Just giving that answer doesn't really help with understanding.
We try, on this site, to help students with problems, not just provide the solution for them.

 

[Deleted member 4993]

okay thank you

You should thank your lucky stars that you got help from super patient (and super knowledgeable) tutor to walk you through.

Don't make her mad - she has a pet Tasmanian Devil ......

 

[Ihatdogs10]

If you know triangular similarities then this is a piece of cake for you!

First join PX and YQ
observe that,
∠YOQ = ∠XOP (vertically opp ∠s) and ∠YQO = ∠XPO = 90°
⟹ΔXPO ~ ΔYQO (by AA similarity)
⟹ PX/QY = PO/QO
⟹PX ⋅ QO = QY ⋅ PO
∴ Proved

Hope it helps!

thank you