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Bedi999

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I used similar triangles. Constructing QTP and proved, alternate angles equal and vert. opp hence third angle equal therefor qtp collinear??? not suree :(
 

J2good4u

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I sat my HSC last year but looking at this question would this be valid:

let angle QBA = x
therfore angle BAP = x (alternate angles in parallel lines...)
Similarly, let angle BQP = y
therfore angle APQ = y (alternate angles in parallel lines...)

Now, angle QTA = x + y ( the exterior triangle BQT is equal to the sum of the opposite interior angles)
Also, angle ATP = 180 - (x + y) ( angle sum of triangle PAT 180)

Therefore angle QTP = x + y + 180 - (x +y)
= 180

Thus Q,T,P are collinear
 

BlugyBlug

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You could assume BTA was a straight line, yes?

Then go on to prove the tangent/alternate segment angle etc to get QTP is a straight line?
The line they gave us was straight i'm pretty sure.
 

Web Addict

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This is how I did it, which I'm pretty sure is right:

This is pretty much what I did. However, I did do the beta part. I just wrote angle QTX = angle PTY = alpha, therefore P, T and Q are collinear because vertically opposite angles are equal. Would this be accepted?
 

SpiralFlex

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This is pretty much what I did. However, I did do the beta part. I just wrote angle QTX = angle PTY = alpha, therefore P, T and Q are collinear because vertically opposite angles are equal. Would this be accepted?
No. [Assuming you said vertically opposite angles first] iBibah got angle QTX as alpha from a different argument. Cannot say vertically opposite because there is no guarantee QTP is straight UNLESS you used a different argument of finding alpha.
 

Web Addict

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No. [Assuming you said vertically opposite angles first] iBibah got angle QTX as alpha from a different argument. Cannot say vertically opposite because there is no guarantee QTP is straight UNLESS you used a different argument of finding alpha.
I didn't use vertically opposite angles first. To get the alphas, I used the alternate segement theroem, then I conclude that P, T and Q are collinear because the vertically opposite angles are equal.
 

CrackerMo

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I sat my HSC last year but looking at this question would this be valid:

let angle QBA = x
therfore angle BAP = x (alternate angles in parallel lines...)
Similarly, let angle BQP = y
therfore angle APQ = y (alternate angles in parallel lines...)

Now, angle QTA = x + y ( the exterior triangle BQT is equal to the sum of the opposite interior angles)
Also, angle ATP = 180 - (x + y) ( angle sum of triangle PAT 180)

Therefore angle QTP = x + y + 180 - (x +y)
= 180

Thus Q,T,P are collinear
That's the way I did it :S
 

Firmin

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I didn't use vertically opposite angles first. To get the alphas, I used the alternate segement theroem, then I conclude that P, T and Q are collinear because the vertically opposite angles are equal.
+1 It's valid because you prove they're equal angles and THEN you state well since they're equal, they must be vertically opposite. Hence straight line.
 

Hotisgood5

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haha, this feels like umat all over again. Just because we derived the vertically opposite rule from straight lines, we cant say because vertically opposite angles are equal, the two lines MUST be straight. The logic is definitely true is we go forward, but we cant guarantee that the logic is also true when we go backwards.
 
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andybandy

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I knew the right answer, but for some reason i kept doing the wrong answer.. ahh that was stupid of me
 

Recondit

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I constructed a common tangent MN through T, and lines QT and TP
Said <QBT=<PAT (alternate angles are equal on parallel lines QB//AP)
Said <QTM=<QBT (alternate segment theorem)
Said <PTN=<PAT (alternate segment theorem)
Therefore, <QTM=<PTN
Since <QTM and <PTN are vertically opposite angles and are equal
Therefore Q, T, P are collinear

???
 

panda15

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I sat my HSC last year but looking at this question would this be valid:

let angle QBA = x
therfore angle BAP = x (alternate angles in parallel lines...)
Similarly, let angle BQP = y
therfore angle APQ = y (alternate angles in parallel lines...)

Now, angle QTA = x + y ( the exterior triangle BQT is equal to the sum of the opposite interior angles)
Also, angle ATP = 180 - (x + y) ( angle sum of triangle PAT 180)

Therefore angle QTP = x + y + 180 - (x +y)
= 180

Thus Q,T,P are collinear
Not valid, because you assumed that QBP was a straight line, which was what the question was asking you to prove.
 

MATHmaster

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Do you still think that deserves 1/3 since the first statement is correct in completing the proof?
 

SpiralFlex

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Do you still think that deserves 1/3 since the first statement is correct in completing the proof?
I would think it would be 1/3 though I am not familiar with the marking guidelines. But 1/3 makes sense.
 

RealiseNothing

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I sat my HSC last year but looking at this question would this be valid:

let angle QBA = x
therfore angle BAP = x (alternate angles in parallel lines...)
Similarly, let angle BQP = y
therfore angle APQ = y (alternate angles in parallel lines...)


Now, angle QTA = x + y ( the exterior triangle BQT is equal to the sum of the opposite interior angles)
Also, angle ATP = 180 - (x + y) ( angle sum of triangle PAT 180)

Therefore angle QTP = x + y + 180 - (x +y)
= 180

Thus Q,T,P are collinear
That's assuming the result, so no.
 

panda15

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Do you still think that deserves 1/3 since the first statement is correct in completing the proof?
Maybe 1/3 for using the alternate angles between parallel lines on the line that was given to you as straight,
 

Firmin

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Construct tangent ZY passing through T and is tangential to C1 & C2
<QTZ=<QBT (angle between tangent and chord is equal to angle subtended by that chord in the alternate segment)
Similarly <YTP = <TAP
<TAP=<QBT (alternate angles on parallel lines are equal QBllPA)
Therefore <QTZ=<YTP (transitivity of equality)
Therefore QTP is a straight line (converse of vertically opposite angles are equal)
 

obliviousninja

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I was such a silly billy, I had no clue how to prove collinearcy, i just proved similar triangles. Lol. Herpa Derp
 

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