Similarity problem, tangents

• Aug 31st 2009, 06:23 AM
HelenaStage
Similarity problem, tangents
Hi,
So I have a problem that reads as follows:

[PQ] is a chord of a circle. R lies on the major arc of the circle. Tangents are drawn through P and through Q. From R, perpendiculars [PA], [PB] and [PC] are drawn to the tangent at A, the tangent at B and [AB] respectively. Prove that RA.RB = RC^2.

It's not that I don't get similarity, it's rather the phrasing of the problem...how can you have perpendiculars PA, PB and PC from R??? (Speechless)
• Aug 31st 2009, 07:03 AM
bowline
Quote:

Originally Posted by HelenaStage
Hi,
So I have a problem that reads as follows:

[PQ] is a chord of a circle. R lies on the major arc of the circle. Tangents are drawn through P and through Q. From R, perpendiculars [PA], [PB] and [PC] are drawn to the tangent at A, the tangent at B and [AB] respectively. Prove that RA.RB = RC^2.

It's not that I don't get similarity, it's rather the phrasing of the problem...how can you have perpendiculars PA from R??? (Speechless)

You can't !

At least not until you define where points A, B and C are.
The problem is incomplete in its current form.
• Aug 31st 2009, 07:08 AM
HelenaStage
Quote:

Originally Posted by bowline
You can't !

At least not until you define where points A, B and C are.
The problem is incomplete in its current form.

That's all the info I get...
• Aug 31st 2009, 07:25 AM
aidan
Quote:

Originally Posted by HelenaStage
Hi,
So I have a problem that reads as follows:

[PQ] is a chord of a circle. R lies on the major arc of the circle. Tangents are drawn through P and through Q. From R, perpendiculars [PA], [PB] and [PC] are drawn to the tangent at A, the tangent at B and [AB] respectively. Prove that RA.RB = RC^2.

It's not that I don't get similarity, it's rather the phrasing of the problem...how can you have perpendiculars PA from R??? (Speechless)

Quote:

the phrasing is the problem

A simple drawing would explain it.
Unfortunately, I do not understand the phrasing well enough to construct a sketch.

You are drawing tangents to A and B, therefore A & B must be on the circle. It does not so state but C can be anywhere.

Quote:

how can you have perpendiculars PA from R???
If D is on the line PA, then line RD is perpendicular to line PA; it is a line perpendicular to PA that passes through point R.
• Aug 31st 2009, 07:31 AM
HelenaStage
Quote:

Originally Posted by aidan
A simple drawing would explain it.
Unfortunately, I do not understand the phrasing well enough to construct a sketch.

You are drawing tangents to A and B, therefore A & B must be on the circle. It does not so state but C can be anywhere.

If D is on the line PA, then line RD is perpendicular to line PA; it is a line perpendicular to PA that passes through point R.

Yes, OK, but the problem says perpendiculars PA, PB and PC...?

I believe point C will be the result of a line from R cutting the line joining A and B...
• Sep 1st 2009, 07:14 AM
aidan
Quote:

Originally Posted by HelenaStage
[PQ] is a chord of a circle. R lies on the major arc of the circle. Tangents are drawn through P and through Q. From R, perpendiculars [PA], [PB] and [PC] are drawn to the tangent at A, the tangent at B and [AB] respectively. Prove that RA.RB = RC^2.

Quote:

That's all the info I get...
See the attached sketches for my initial understanding of the question.
• Sep 1st 2009, 07:18 AM
aidan
Since I'm limited to attaching 5 files
this is a continuation of the above: