How to Understand Basic Geometry Theorems for Circles?

[disregardthat]

My problem is how to understand this problem. I am confident that I can solve it, but I need to know what they mean in the problem.

Homework Statement

"[PQ] is a chord of a circle. R lies on the major arc of the circle."
I understand this.

"Tangents are drawn through P and through Q."
What does this mean? I would say that they meant that there are tangents touching the circle in P and Q.

"From R, perpendicuars [PA], [PB] and [PC] are drawn to the tangent at A, the tangent at B and [AB] respectively."
I'm totally lost. What do they mean with this?

"Prove that $$RA.RB=RC^2$$"
This is the problem I am to prove, when I understand the problem.

Homework Equations

Basic geometry theorems

The Attempt at a Solution

Just need some help with understanding the problem.
Thanks

 

[JoshG]

Did this come with a diagram, by any chance?

 

[disregardthat]

Nope. In problems with a diagram, the text "In the given figure" appears.

 

[JoshG]

Something just doesn't make sense to me.

"From R, perpendicuars [PA], [PB] and [PC] are drawn to the tangent at A, the tangent at B and [AB] respectively."

Since the perpendiculars are labeled PA, PB, and PC, I'm assuming that they all originate from point P. So how could they be drawn from R? I'm also assuming that points A and B lie on the circle. If so, the lines perpendicular to them would have to go through the center, and it should not be possible to draw lines perpendicular to two different tangents from the same point on the circle, whether R or P.

Only one perpendicular to a tangent can be drawn from R, and that's the diameter through R. The same goes for P, since P also lies on the circle.

I can't help but think there's some missing information here. I'm sorry if this just adds to the confusion, but I just can't see how any figure could possibly be drawn to the specifications here.

 

[disregardthat]

Since the perpendiculars are labeled PA, PB, and PC, I'm assuming that they all originate from point P. So how could they be drawn from R? I'm also assuming that points A and B lie on the circle. If so, the lines perpendicular to them would have to go through the center, and it should not be possible to draw lines perpendicular to two different tangents from the same point on the circle, whether R or P.

Excactly! :/

I can't either see the figure. I guess I could skip this excersise with good conscience then :)

Thanks

(I think this chapter in the book was hastily written together)

 

[HallsofIvy]

My problem is how to understand this problem. I am confident that I can solve it, but I need to know what they mean in the problem.

Homework Statement

"[PQ] is a chord of a circle. R lies on the major arc of the circle."
I understand this.

"Tangents are drawn through P and through Q."
What does this mean? I would say that they meant that there are tangents touching the circle in P and Q.

Yes, since there is only one circles, the tangents are tangents to the circle at P and Q.

"From R, perpendicuars [PA], [PB] and [PC] are drawn to the tangent at A, the tangent at B and [AB] respectively."
I'm totally lost. What do they mean with this?

Since it says "from R", I feel sure the "P" is a type. From the point R, you draw a perpendicular to the tangent through P: call the foot of the perpendicular A. From the point R, you draw a perpendicular to the tangent through Q: call the foot of the perpendicular B. Also draw line segment AB. Finally construct the perpendicular from R to setment AB. Call the foot of that C.

"Prove that $$RA.RB=RC^2$$"
This is the problem I am to prove, when I understand the problem.

Homework Equations

What you are to prove is that the distance from R to A, times the distance from R to B, is the square of the distance from R to C, squared.

Basic geometry theorems

The Attempt at a Solution

Just need some help with understanding the problem.
Thanks

 

[disregardthat]

Thanks for the reply! It is no way that C is on AB though. RC squared cannot be the same as the product of RA-RB. I drew the figure in geogebra, and if we place the perpendicular line from R to the segment PQ, then the thing we are to prove seem very correct. I try to move the point R around on the figure, and RA.RB-RC^2 = 0 for all points R on the circle. I will prove this instead. If C were on AB, it would not be right for any R, as RC is smaller than both RA and RB for any R (on my figure).


Proof: (See attached picture for helping figure)

Let the tangents in P and Q intersect in M. Let $$\angle MPQ= \alpha$$. As $$\Delta MPQ$$ is an isosceles, $$\angle PQM = \alpha$$. By the "angle between a chord and a tangent" theorem, $$\angle PRQ = \alpha$$.
Let $$\angle ARP = \beta \Rightarrow \angle RPA = 90-\beta \Rightarrow \angle RPC = 90+\beta - \alpha$$.
Now in $$\Delta PRQ$$, $$\angle PRQ - \angle CRQ + \angle RPC +\angle PCR = 180 \Rightarrow \angle CRQ = \beta$$.

Now we have that $$\Delta ARP$$ is similar to $$\Delta CRQ$$.
Then $$\frac{RA}{RP}=\frac{RC}{RQ} \Rightarrow \frac{RC}{RA}=\frac{RQ}{RP}$$.

Now as $$\angle CQR = 90 - \beta, \angle RQB=90+\beta - \alpha \Rightarrow \Delta PCR$$ is similar to $$\Delta QBR$$.
Then $$\frac{RP}{RC}=\frac{RQ}{RB} \Rightarrow \frac{RQ}{RP}=\frac{RB}{RC}$$ and hence is $$\frac{RC}{RA}=\frac{RB}{RC} \Rightarrow RA.RB=RC^2$$

$$Q.E.D$$

Does it seem ok?

I believe that this is what the problem meant.

 

[belliott4488]

Thanks for the reply! It is no way that C is on AB though. RC squared cannot be the same as the product of RA-RB. I drew the figure in geogebra, and if we place the perpendicular line from R to the segment PQ, then the thing we are to prove seem very correct. I try to move the point R around on the figure, and RA.RB-RC^2 = 0 for all points R on the circle. I will prove this instead. If C were on AB, it would not be right for any R, as RC is smaller than both RA and RB for any R (on my figure).


Proof: (See attached picture for helping figure)

Let the tangents in P and Q intersect in M. Let $$\angle MPQ= \alpha$$. As $$\Delta MPQ$$ is an isosceles, $$\angle PQM = \alpha$$. By the "angle between a chord and a tangent" theorem, $$\angle PRC = \alpha$$.

I think this was a typo. You meant $$\angle PRQ = \alpha$$, right?

Let $$\angle ARP = \beta \Rightarrow \angle RPA = 90-\beta \Rightarrow \angle RPC = 90+\beta - \alpha$$.
Now in $$\Delta PRQ$$, $$\angle PRQ - \angle CRQ + \angle RPC +\angle PCR = 180 \Rightarrow \angle CRQ = \beta$$.

Now we have that $$\Delta ARP$$ is similar to $$\Delta CRQ$$.
Then $$\frac{RA}{RP}=\frac{RC}{RQ} \Rightarrow \frac{RC}{RA}=\frac{RQ}{RP}$$.

Now as $$\angle CQR = 90 - \beta, \angle RQB=90+\beta - \alpha \Rightarrow \Delta PCR$$ is similar to $$\Delta QBR$$.
Then $$\frac{RP}{RC}=\frac{RQ}{RB} \Rightarrow \frac{RQ}{RP}=\frac{RB}{RC}$$ and hence is $$\frac{RC}{RA}=\frac{RB}{RC} \Rightarrow RA.RB=RC^2$$

$$Q.E.D$$

Does it seem ok?

I believe that this is what the problem meant.

Looks good to me! Well done.

I had also initially thought that the point C was supposed to be on the the line AB, but your proof shows that it must be as you interpreted it.

 

[disregardthat]

Thanks,
that problem had a couple of typos.

yeah, I meant that.