+1124 Hi everyone!,
I am working through a paper, and have completed so to speak the majority of problems. I got stuck with another really tricky one. Please if you could, how can I prove "O" is the centre?
GIVEN:
1) AOIIQR
2) PA = AQ
3) PB = BT
First I had to prove ABIIQT, which I did. I just went by saying:
In triangle PQT:
PA=AQ Given
PB=BT Given
therefore AB disects two sides of triangle
therefore ABIIQT ...
NOW, prove "O" is the centre of the circle if PR is the center line of the circle.
All I do is stare at this...I have thought about the following however:
Since AO runs from the centre of PQ, parallel to the 3rd side (QR), does it not mean that AO will disect PR into two exact halves?, and since PR runs from one side of the circle to the other side of it, it would mean that PO is equal to OR...which tells me "O" is the centre?....I'm not sure if I'm at all close here...Please advise? Thank you all..
' Since AO runs from the centre of PQ, parallel to the 3rd side (QP), does it not mean that AO will disect PR into two exact halves ' ?
Yes it does, but this is what you are required to prove.
Show that the two triangles PAO and PQR are similar and then consider the ratio PA/PO.
Equate it with the corresponding ratio in the larger triangle and then use the property that PA = AQ.
+1124 Hi guest,
thank you very much for nudging me into the right direction...I should be able to get this now...Thanx a lot!
+1124 Hi Melody,
well, honestly....I went about the whole thing a different way...The reason I believe I cannot make use of ratio's, is because I do not have any values. I do not know how one can use ratio's if there are no values....so I did this:
In triangle PQR:
AOIIQR Given
PA=AQ Given
Therefore AO runs through the middel of triangle PQR
Therefore AO devides PR Midpoint theorem
Thus PO=OR
Since PR is the diameter, Given
PO must equal OR
Thus "O" is the middle of the circle.
Is this wrong?..Thanx for your assistance!!
+128407
Since AO is ll to QR then PR is a transversal that cuts parallel lines
Thus angle POA = angle PRQ
And angle QPR = angle APO
Thus, by Angle-Angle congruency....triangle QPR is similar to triangle APO
But similar triangles are similar in all respects.....
So........ PQ / PR = PA / PO which also implies that PA / PQ = PO / PR
But.....PA = AQ.....so.....PA = (1/2)PQ
Then....by substitution.......
(1/2) PQ / PQ = PO / PR
Simplifying.....we have that
(1/2) = PO / PR whicih implies that
PO = (1/2)PR
And since PR is the diameter of the circle and PO is (1/2) of PR.....O must be the center of the circle
+1124 CPhill,
I really wish I had the experience you guys have!!..It's good to see how it's solved, however, please if you do not mind....what in my solution makes it wrong?..I understand my approach was wrong, simply because you say it was, and you gave me the appropriate solution....but why is my approach wrong?..would you kindly spend just a little more time please, and just educate me..please...I need to understand this?..You guys are great, thank you for being out there!!
+118608 Hi Chris,
I followed your logic right up till you said "And since PR is the diameter of the circle"
Why can you say that PR is the diameter of the circle ??
Maybe it is written somewhere and i am blind. I am not being sarcastic, this is always a distinct posibility :)
It is Red's fault! He stole my glasses!
+118608 Hi Juriemagic,
Firstly you disect rats in a laboratory and you bisect intervals when you cut them in half :)
Yes I like your logic, although you would need to discuss similar triangles as Chris did.
I have the same problem with your logic as I do with Chris's, I can see that O bisects PR easily enough but I do not know why either of you has stated that PR is a diameter 
I now see that the original guest answerer has made exactly the same 'assumption' ://
+1124 haha...disects!!..oh my goodness!!!..so sorry, that was an honest boo boo... 
Also, This exact question I found on the inernet in an exam question paper. The marks allocated for proving ABIIQT, was 2, and for proving the centre, also just 2. Sooo, I'm thinking going the similar triangle way is a lot of work for just 2 marks, HOWEVER, I do aknowledge that the approach given to me was far superior and most likely, the best approach.
Guys, you have all been a great help with this one, I really admire and love you all!!!
+118608 LOL That is what I thought the first time I saw it but you did that boo boo a couple of times :)
Did you put your working in the middle of the question as I have asked further down the page?
Please be careful. It is very frustrating to put a lot of time into a question when the question is not presented properly. :/
No assumption Melody.
Read the question.
NOW, prove "O" is the centre of the circle if PR is the centre line of the circle.
+118608 I thought that statement was a part of Juriemagic's working.
I did not think it was a part of the question!
------------------
This is the the question, along with the pic of course:
I am working through a paper, and have completed so to speak the majority of problems. I got stuck with another really tricky one. Please if you could, how can I prove "O" is the centre?
GIVEN:
1) AOIIQR
2) PA = AQ
3) PB = BT
No where does it say that PR is a diameter!
-------------------
NOW that I look at it with your eyes I can see that maybe Juriemagic put his working in the middle of his question.
All very confusing!
+1124 Okay, let me clarify this...
The paper reads:
GIVEN:
1) AOIIQR
2) PA = AQ
3) PB = BT
Prove:
1) ABIIQT
2) O is the centre of the circle if PR is the centre line through the circle
3) BORT is a trapezium
Sorry if there was confusion, I did not realize giving it like it actually was on the paper would or could cause a mis-understanding. From my side, please accept my apologies.
+118608 ok Juiemagic but you left out key components of the question and the other question I spent a lot of time on also had bits missing.
Last time I thought it was a badly written, incomplete question but maybe the question was just fine until you gave us the condensed version.
I certainly accept your appology but please do not give us half questions (or reworded questions) any more. :/
