Circle Geometry (1 Viewer)

taggs-sasuke · May 20, 2009

Hello...

I have no idea how to prove the part in BOLD:

O and C are the centres of two circles which intersect at A and B. (AB is the common chord.)

Prove triangleOAC is congruent to triangleOBC

Prove angleAOZ is equal to angleBOZ

and hence prove that the line of centres OC bisects the common chord AB at right angles.

Can someone please help me?

I'd be really grateful.

Thanks in advance.

I have ANOTHER question.

Two concentric circles whose common centre is O are cut by a third circle whose centre is C. The common chords are AB and PQ. Prove AB is parallel to PQ.

Thanks

kurt.physics · May 20, 2009

taggs-sasuke said:
Hello...

I have no idea how to prove the part in BOLD:

O and C are the centres of two circles which intersect at A and B. (AB is the common chord.)

Prove triangleOAC is congruent to triangleOBC

Prove angleAOZ is equal to angleBOZ

and hence prove that the line of centres OC bisects the common chord AB at right angles.

Can someone please help me?

I'd be really grateful.

Thanks in advance.

I have ANOTHER question.

Two concentric circles whose common centre is O are cut by a third circle whose centre is C. The common chords are AB and PQ. Prove AB is parallel to PQ.

Thanks

Is Z where AB intersects with OC?

kurt.physics · May 20, 2009

I will assume that it is:

$We have:$

$\triangle OAC \cong \triangle OBC$

$\therefore \angle AOZ = \angle BOZ\,\, $(corresponding angles in congruent triangles)$$

$AO = BO\,\,$(equal radii)$$

$$OZ is common to both$

$By SAS,$

$\triangle AOZ \cong \triangle BOZ$

$\therefore \angle AZP = \angle BZO\,\,$(corresponding angles in congruent triangles)$$

$But$\,\,\angle AZO + \angle BZO = 180\^{\circ}\,\,$(Angles on a straight line)$

$\therefore \angle AZO = \angle BZO = 90^{\circ}$

$\therefore\,\,$OC bisects AB at right angles$$

kurt.physics · May 20, 2009

taggs-sasuke said:
Hello...

I have ANOTHER question.

Two concentric circles whose common centre is O are cut by a third circle whose centre is C. The common chords are AB and PQ. Prove AB is parallel to PQ.

Thanks

I will just do this one quick, if you dont understand im happy to elaborate.

$Construct the radii PO and QO$

$AO = BO $(equal radii)$$

$PO = QO $(equal radii)$$

$\therefore \frac{AO}{PO} = \frac{BO}{QO}$

$\angle AOB = \angle POQ\,\,$(common angle)$$

$\therefore \triangle AOB \parallel \triangle POQ \,\,$(SAS)$$

$\therefore \angle OAB = \angle OPQ\,\,$(corresponding angles in similar triangles)$$

$\therefore PQ \parallel AB\,\,$(corresponding angles)$$

*note: when i wrote the similaritys of the triangles (by SAS) i used the parallel symbol because i didnt know the code for the symbol for similarity
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taggs-sasuke · May 20, 2009

kurt.physics said:
Is Z where AB intersects with OC?

Yes, it is.

Sorry.

But yay, you correctly assumed it is.

taggs-sasuke · May 20, 2009

kurt.physics said:
I will assume that it is:

$We have:$

$\triangle OAC \cong \triangle OBC$

$\therefore \angle AOZ = \angle BOZ\,\, $(corresponding angles in congruent triangles)$$

$AO = BO\,\,$(equal radii)$$

$$OZ is common to both$

$By SAS,$

$\triangle AOZ \cong \triangle BOZ$

$\therefore \angle AZP = \angle BZO\,\,$(corresponding angles in congruent triangles)$$

$But$\,\,\angle AZO + \angle BZO = 180\^{\circ}\,\,$(Angles on a straight line)$

$\therefore \angle AZO = \angle BZO = 90^{\circ}$

$\therefore\,\,$OC bisects AB at right angles$$

So we have to prove that angleAZO and angleBZO are right angles but we don't prove that AZ = BZ?

taggs-sasuke · May 20, 2009

kurt.physics said:
I will just do this one quick, if you dont understand im happy to elaborate.

$Construct the radii PO and PQ$

$AO = BO $(equal radii)$$

$PO = QO $(equal radii)$$

$\therefore \frac{AO}{PO} = \frac{BO}{QO}$

$\angle AOB = \angle POQ$

$\therefore \triangle AOB \parallel \triangle POQ \,\,$(SAS)$$

$\therefore \angle OAB = \OPQ\,\,$(corresponding angles in similar triangles)$$

$\therefore PQ \parallel AB$

*note: when i wrote the similaritys of the triangles (by SAS) i used the parallel symbol because i didnt know the code for the symbol for similarity

No, I do understand. Thanks.

P.S. There are a few typos (for yourself kurt.physics and those who are viewing this solution).

kurt.physics · May 20, 2009

taggs-sasuke said:
So we have to prove that angleAZO and angleBZO are right angles but we don't prove that AZ = BZ?

Oh, yeh, sorry. The last part is just

$AZ = BZ\,\,$(corresponding sides of congruent triangles)$$

gurmies · May 20, 2009

kurt.physics, you love your circle geometry

kurt.physics · May 20, 2009

taggs-sasuke said:
No, I do understand. Thanks.

P.S. There are a few typos (for yourself kurt.physics and those who are viewing this solution).

I just read over it, there are so many! I fixed them up by the way. Sorry for that, it must have made it really hard to understand :S I had to write it really really fast because i had to go to dinner

kurt.physics · May 20, 2009

gurmies said:
kurt.physics, you love your circle geometry

I sure do!

gurmies · May 20, 2009

To be honest I do as well, except I hate drawing the diagrams myself. Provide a diagram for me and I think I can do most of them xD

kurt.physics · May 20, 2009

gurmies said:
To be honest I do as well, except I hate drawing the diagrams myself. Provide a diagram for me and I think I can do most of them xD

yeh, just getting the graph right solves half the problem!

gurmies · May 20, 2009

Bingo.

Gibbatron · May 20, 2009

It is a pain in the arse. There are so many ways to find each solution, and it gets confusing :S

gurmies · May 20, 2009

When in doubt, angle in the alternate segment

taggs-sasuke · May 20, 2009

kurt.physics said:
I just read over it, there are so many! I fixed them up by the way. Sorry for that, it must have made it really hard to understand :S I had to write it really really fast because i had to go to dinner

Haha.

Don't worry, it wasn't.

taggs-sasuke · May 20, 2009

kurt.physics said:
Oh, yeh, sorry. The last part is just

$AZ = BZ\,\,$(corresponding sides of congruent triangles)$$

Ok, thanks.

Woteva636 · May 20, 2009

Gibbatron said:
It is a pain in the arse. There are so many ways to find each solution, and it gets confusing :S

Yeah, its like the more you know the less you know, because there's more ways to do it.

taggs-sasuke · May 20, 2009

Woteva636 said:
Yeah, its like the more you know the less you know, because there's more ways to do it.

I've only begun learning it and I dislike it already.

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