[QUOTE=mfgoode][QUOTE=Chris Card]I didn't quite understand mfgoode's explanation, it is probably correct, but my solution is:
[spoiler]
Suppose the point P is on the line AC, and is such that the length PC is a fraction x of the length AC, with 1/2 <= x <= 1 (relabel the triangle as necessary to satisfy these conditions. Then if Q is the point on the line BC such that PQ divides the triangle's area in half, then the length QC is a fraction y of the length BC, where xy = 1/2.
The only problem is that I don't know how (or if it's possible) to construct the point Q - anyone?
[/spoiler]

Chris[/QUOTE
smile:
Your solution is really a question as you have not solved anything rather confused yourself further.
Kindly read my solution again.
Without the aid of a diagram your problem seems to be that you must realise that you must get two similar triangles whatever the triangle you start with.
[/QUOTE]
How can you get two similar triangles in general when you cut a triangle with a line? In most cases you get a triangle and a quadrilateral.
[QUOTE]Therefore PQ must be parallel to the side it does not intersect. Dont worry and waste time to where point Q will intersect BC. The // line PQ does the trick.
[/QUOTE]
A line parallel to the side it does not intersect will not divide the area of the triangle in half, except for some special cases. Either your statement is wrong, or I'm not understanding what you are saying, which is certainly possible.
[QUOTE]
The ratio you have used is wrong. Go over the posts. Every post is important including the one on COG. The theorem is clearly and precisely stated more than once. You have used the wrong ratio and there fore not understood a gist of the solution because of lack of study of the solution.
You require more effort on your part and I cannot and refuse to spoon feed you any further.
And you have the audacity to state that my solution is 'probably correct'?
[/QUOTE]
No, I couldn't understand your explanation, and gave you the benefit of the doubt.
[QUOTE]
And you introduced a hairbrained solution of your own when you dont know the basic principles and expect to fox others with it?
Mally :coffee[/QUOTE]
Given a triangle ABC, with length AC = a, length BC = b, angle at C = gamma,
the area of the triangle ABC is 1/2 ab sin(gamma). If point P is on AC, and point Q is on BC, length PC = p, length QC = q, then by the same formula, area of the triangle PCQ is 1/2 pq sin(gamma).
[spoiler]So, if PQ splits the triangle into 2 halves, we must have pq = 1/2 ab.
If p = l a, q = m b, we have lm = 1/2.[/spoiler]

Chris

And this is how you construct the point Q:
[spoiler]
We have a triangle ABC with a point P on line AC (say P is nearer to A than C for this argument). Construct the line from P to
B and then construct the line parallel to this line which passes through the mid-point of AC. The point Q is where the latter line intersects the line BC.
[/spoiler]

Chris

Proof
 

[QUOTE=Chris Card]And this is how you construct the point Q:
[spoiler]
We have a triangle ABC with a point P on line AC (say P is nearer to A than C for this argument). Construct the line from P to
B and then construct the line parallel to this line which passes through the mid-point of AC. The point Q is where the latter line intersects the line BC.
[/spoiler]

Chris[/QUOTE]
:rolleyes: Excellent Chris Card. Im wondering how to prove it. Could you help out?
Mally :coffee:

[QUOTE=mfgoode]:rolleyes: Excellent Chris Card. Im wondering how to prove it. Could you help out?
Mally :coffee:[/QUOTE]
[spoiler]
Yes I can prove it. Just use the formula I gave before for the area of a triangle in terms of 2 sides and the angle between them. All you have to do is
work out the length of the line QC and then apply the formula to the triangle PQC.
To do that, notice that the triangle PBC is similar to RQC where R is the midpoint of AC.
So if p is the length PC, q is the length QC, a is the length AC and b is the length BC, we have
p/(1/2 a) = b / q, or pq = 1/2 ab.
[/spoiler]

[QUOTE=Chris Card][spoiler]
Yes I can prove it. Just use the formula I gave before for the area of a triangle in terms of 2 sides and the angle between them. All you have to do is
work out the length of the line QC and then apply the formula to the triangle PQC.
To do that, notice that the triangle PBC is similar to RQC where R is the midpoint of AC.
So if p is the length PC, q is the length QC, a is the length AC and b is the length BC, we have
p/(1/2 a) = b / q, or pq = 1/2 ab.
[/spoiler][/QUOTE]

:smile: Thanks Chris Card. Thats a way out I suppose but not concise enough neither elegant in my opinion if I may dare say so.

Of hand I do not think there is any necessity to work out areas and lengths of sides which makes the proof unnecesarily long and tiresome.

Im sure you should make use of the median belonging to the centrepoint you have mentioned which divides the given triangle in half.
Then prove the triangle you have constructed to be equal to it as it is half the given triangle.
If there is any difficulty I am willing to give the full proof as I see it.

Mally :coffee:

[QUOTE=mfgoode]:smile: Thanks Chris Card. Thats a way out I suppose but not concise enough neither elegant in my opinion if I may dare say so.

Of hand I do not think there is any necessity to work out areas and lengths of sides which makes the proof unnecesarily long and tiresome.

Im sure you should make use of the median belonging to the centrepoint you have mentioned which divides the given triangle in half.
Then prove the triangle you have constructed to be equal to it as it is half the given triangle.
If there is any difficulty I am willing to give the full proof as I see it.

Mally :coffee:[/QUOTE]
I'm sure the proof can be stated more elegantly, but this method also allows a more general result to be easily proved : finding a point Q such that area PQC is any fraction of area ABC.

Chris

PR#25
 

[QUOTE=Chris Card]I'm sure the proof can be stated more elegantly, but this method also allows a more general result to be easily proved : finding a point Q such that area PQC is any fraction of area ABC.

Chris[/QUOTE]
:smile:

Not really and no problem. Just generalise it to any ratio.

Well Chris sorry for the delay in replying. World Cup fever you know.

I present a simpler proof than yours and hope you will agree with me.

The construction is the same as yours but my lettering is different for my convenience in presenting it.

TO PROVE:
P is a fixed point on any side of any triangle ABC.

Thru it draw a line PQ meeting a side at Q such the Area of the triangle formed is in a ratio 1/r of the Area of the triangle ABC.

CONSTRUCTION:
:
Draw triangle ABC lettering it anti - clockwise such that BC is the base and A is the vertex.

Let P be the given fixed point but lying on BC nearer to C.

Join A to P

Divide BC in the ratio (1/r) required, at D. [ In the previous case this was ½.]

Thru D draw line parallel (//) to AP meeting side AB in Q.

Join A to D and Q to P.
Hence Q is the reqd. point and BPQ is the reqd. triangle.

PROOF:

On line QD as base the triangles (tng.) DQA = (tng)) DPQ
[(tng.) Abbr. for triangle]

Reason?, (tng.’s) on same base and between the same (//’s) QD and AP. [Theorem]

Now add to each (tng.) the (tng.) BDQ

Thus we get Area (tng) BPQ = Area (tng.) ABD.

Now Area (tng) ABD is (1/r) Area (tng) ABC the given triangle. Because

Base BD is (1/r) base BC [by construction]

Therefore Area (tng.) BPQ = (1/r) Area (tng.) ABC.
Q.E.D.

Mally. :coffee:

P.S. Thanks for the insight to generalise.