# Prove PR = QS

#### Albert

##### Well-known member
HI : good to see you everybody ,I am a new comer on this board
Being a math teacher ,I always trained my students with various difficult problems
from now on I am going to post a sequential challenging questions for people to share
most of them I know the answer and the solution of it , maybe your solution is quicker and smarter

here comes the problem :

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#### Sudharaka

##### Well-known member
MHB Math Helper
Re: Prove PR=QS

Hi Albert,

The second sentence in your question isn't clear to me.

a line passing through point B and intersects two points $$P,\,Q$$ with $$O$$ and intersects $$R,\, S$$ with $$O_1$$ and $$O_2$$, respectively prove: $$\overline{PR} = \overline{QS}$$
There are infinitely many lines passing though $$B$$ and not all of them will have $$PR = QS$$. Can you please elaborate a bit more on what you mean by this sentence.

Kind Regards,
Sudharaka.

#### Albert

##### Well-known member
Re: Prove PR=QS

Can you plot a line passing through point B ,
satisfying the restriction and show that PR and QS are not equal ?

#### Sudharaka

##### Well-known member
MHB Math Helper
Re: Prove PR=QS

Can you plot a line passing through point B ,
satisfying the restriction and show that PR and QS are not equal ?
I might be missing something here but what is meant by "the restriction"? Do you mean that $$O$$ lies on the perpendicular bisector of $$PQ$$ ?

#### Albert

##### Well-known member
Re: Prove PR=QS

I might be missing something here but what is meant by "the restriction"? Do you mean that $$O$$ lies on the perpendicular bisector of $$PQ$$ ?
yes, that must be

#### Sudharaka

##### Well-known member
MHB Math Helper
Re: Prove PR=QS

yes, that must be
But in that case there are infinitely many lines as I have claimed above, since for every chord of a circle the perpendicular bisector passes through the center of the circle. Refer >>this<<.

#### Albert

##### Well-known member
Re: Prove PR=QS

the line passing through point B must also intersect with those of two small circles :
O1,and O2, at point R and S respectively as mentioned before

#### Sudharaka

##### Well-known member
MHB Math Helper
Re: Prove PR=QS

the line passing through point B must also intersect with those of two small circles :
O1,and O2, at point R and S respectively as mentioned before
I think we are misunderstanding each other. To clarify further I have created this diagram using Geogebra and drawn two possible lines for $$PQ$$.

#### Albert

##### Well-known member
Re: Prove PR=QS

where is your points R1,S1 and R2,S2
we want to prove P1R1=Q1S1 , and P2R2=Q2S2

#### Sudharaka

##### Well-known member
MHB Math Helper
Re: Prove PR=QS

I understand the question now. It seems that for all lines drawn through $$B$$, $$PR=QS$$. Sorry for the confusion.

I haven't found the proof yet, but I am trying....

#### Opalg

##### MHB Oldtimer
Staff member
Re: Prove PR=QS

In the diagram, all the angles marked as right angles are angles in a semicircle (and therefore are indeed right angles). The two angles marked $\alpha$ are equal, as are the two angles marked $\beta$ (angles in the same segment). From the triangle $APC$, $AP = AC\cos\alpha$. From the triangle $ARP$, $PR = AP\cos\beta$. So $PR = AC\cos\alpha\cos\beta.$

In the same way, $CQ = AC\cos\beta$ (from triangle $AQC$) and $SQ = CQ\cos\alpha$ (from triangle $CSQ$), and so $SQ = AC\cos\alpha\cos\beta = PR.$

#### Albert

##### Well-known member
Re: Prove PR=QS

Opalg:very good solution

#### Albert

##### Well-known member
Re: Prove PR=QS

here is my solution :