Ratios Of Lines Within Quadrilateral (Proof): Show AX:XC is same as AB:BC

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JSITD

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(a) In a quadrilateral ABCD, the sides AB and CD are parallel, and the diagonal BD bisects angle ABC. Let X be the point of intersection of the diagonals AC and BD. Show that the ratio AX:XC is the same as the ratio AB:BC.

There is also a follow up question, which I'm assuming requires the first question to be solved before it can be attempted. However, it may help to determine the best strategy, so here it is:

(b) In a triangle PQR, the length of each of the three sides is a positive integer. Point M lies on the side QR so that PM is the internal bisector of angle QPR. Also, QM=2 and MR=3. What are the possible lengths of the sides of triangle PQR?

Any help with either or both parts would be much appreciated.
 
Please read the forum guidelines. This is a tutoring web site. We would like to know what you already understand about this exercise, followed by showing whatever attempts you've made. Where exactly are you getting stuck? :cool:
 
I haven't got a lot so far, so I'm looking for some help to get started. I've established that the quadrilateral is a trapeizium (UK) or trapezoid (US). (The definitions are different.) I'm thinking about the shape in terms of triangles ABX and BCX. Line BX is shared between them. Lines AX and XC are collinear. Based on these last two sentences, it should be clear that the ratios are equivalent, but I just can't find a way to express why. I think that it has something to do with the sine rule, but I can't seem to figure out exactly what.
 
JSITD said:
I haven't got a lot so far, so I'm looking for some help to get started. I've established that the quadrilateral is a trapeizium (UK) or trapezoid (US). (The definitions are different.) I'm thinking about the shape in terms of triangles ABX and BCX. Line BX is shared between them. Lines AX and XC are collinear. Based on these last two sentences, it should be clear that the ratios are equivalent, but I just can't find a way to express why.
Click to expand...
Why would "it...be clear", based only on this?

JSITD said:
(b) In a triangle PQR, the length of each of the three sides is a positive integer. Point M lies on the side QR so that PM is the internal bisector of angle QPR. Also, QM=2 and MR=3. What are the possible lengths of the sides of triangle PQR?
Click to expand...
What have you done with this exercise?

Please be complete. Thank you! ;)
 
To answer your first question, I'm not entirely sure why it should be clear (that's the part that I'm struggling with), but I just 'feel' as if it should. I know that that's very non-mathematical, but for some reason I think that collinearity, a shared side and an equal angle should be enough to prove the required ratio. I'm just unclear on what relationship/law to use. :confused:

To answer your second question, part (b) was presented as a follow-up to part (a), in that (a) should be completed before (b) is attempted. This implied to me that I had to complete (a) first and then apply my answer to (b), but if it's possible to do (b) first, then I can definitely attempt it.
 
JSITD said:
(a) In a quadrilateral ABCD, the sides AB and CD are parallel, and the diagonal BD bisects angle ABC. Let X be the point of intersection of the diagonals AC and BD. Show that the ratio AX:XC is the same as the ratio AB:BC.
Click to expand...

JSITD said:
I haven't got a lot so far, so I'm looking for some help to get started. I've established that the quadrilateral is a trapeizium (UK) or trapezoid (US). (The definitions are different.) I'm thinking about the shape in terms of triangles ABX and BCX. Line BX is shared between them. Lines AX and XC are collinear. Based on these last two sentences, it should be clear that the ratios are equivalent, but I just can't find a way to express why. I think that it has something to do with the sine rule, but I can't seem to figure out exactly what.
Click to expand...

JSITD said:
To answer your first question, I'm not entirely sure why it should be clear (that's the part that I'm struggling with), but I just 'feel' as if it should. I know that that's very non-mathematical, but for some reason I think that collinearity, a shared side and an equal angle should be enough to prove the required ratio. I'm just unclear on what relationship/law to use.
Click to expand...

In fact, there is a theorem you may have learned, from which part (a) follows immediately, thinking only of triangle ABC and angle bisector BE. If you don't know it as a theorem, there is a classic way to prove it, using the areas of the triangles ABX and BCX, just as you were thinking. It may also be easy to prove it using the law of sines.

I haven't looked at part (b) yet, but I imagine the same theorem will be used there.
 
I understand the basic shape of the the first shape.. its a parallelogram. with a parallelogram, you know if you were to connect both opposite points that the lines would be equal length. But IDK how to show this mathematically.

The second shape is two triangles. Line QR has a length of 5. IDK if it is a right triangle or if the line MP has a right angle so I am not sure how to solve it from there.
 
Quick said:
I understand the basic shape of the the first shape.. its a parallelogram. with a parallelogram, you know if you were to connect both opposite points that the lines would be equal length. But IDK how to show this mathematically.

The second shape is two triangles. Line QR has a length of 5. IDK if it is a right triangle or if the line MP has a right angle so I am not sure how to solve it from there.
Click to expand...

Quick, are you talking about a whole different problem? It's a trapezoid, not a parallelogram. And there are no equal sides anywhere.

And you can't assume PQR is a right triangle; you can only use what we are told. So whether it is or not is irrelevant.

Let's see what JSITD comes up with, using the ideas that have been discussed, which are sufficient to solve both problems.
 
I've established that part (a) is essentially asking me to prove the angle-bisector theorem. (Something which I haven't actually covered in class, but have just now done some research on.) I was able to do this quite easily using the sine rule.

As far as part (b) is concerned, the same theorem can be used to establish that QR=5, PQ=2n, PR=3n, where n>1. I'm unsure as to whether or not there are any other limits to these possible values, as I can't think of any. I'd appreciate some confirmation on this, however.
 
JSITD said:
I've established that part (a) is essentially asking me to prove the angle-bisector theorem. (Something which I haven't actually covered in class, but have just now done some research on.) I was able to do this quite easily using the sine rule.

As far as part (b) is concerned, the same theorem can be used to establish that QR=5, PQ=2n, PR=3n, where n>1. I'm unsure as to whether or not there are any other limits to these possible values, as I can't think of any. I'd appreciate some confirmation on this, however.
Click to expand...

You have two sides that are in the ratio 2:3, and another side that is 5. The additional restriction will be the triangle inequality: the sum of any two sides must be greater than the third side. That will prevent n from being too large. In fact, you could just try particular pairs (2,3; 4,6; 6,9; ...) to get a feel for what will happen. (I often find that trying specific cases helps in setting up an equation or inequality.)
 
So, bearing in mind the 2:3 ratio, the fact that the sides are all positive integers, and the triangle inequality, the possibilities are: 5,4,6; 5,6,9; 5,8,12. (QR, PQ, PR.)
 
JSITD said:
So, bearing in mind the 2:3 ratio, the fact that the sides are all positive integers, and the triangle inequality, the possibilities are: 5,4,6; 5,6,9; 5,8,12. (QR, PQ, PR.)
Click to expand...

Right: because 5,10,15 or higher is not a triangle.

Or, in terms of an inequality, QR + PQ > PR means 5 + 2k > 3k, so k < 5.
 
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