Hi people, anyone able to solve this problem? I am stumped by (b). Please post the explanation:
http://img258.imageshack.us/img258/6862/problemwx2.th.jpg
The answer for (a) is 5/3
The answer for (b) is 1/5 (Why?)
Sorry, posted the wrong link…
Should be
Homework problem? We don’t solve those here, as a general rule. Can you provide some context, please? 
We may not do homework problems here, but if you’re struggling with that in the wee hours Sunday morning, we can at least tell you that (b) is not 1/5.
Haha not exactly a homework problem since I have graduated more than 10 years ago!
My friend asked me this problem and I am really stumped… if it is not 1/5, what is it???
Eh, a guest saying that doesn’t have a ton of weight.
What information would you need if you were to find the area?
It’s been over two decades since I had any geometry classes, so please refresh my memory: Don’t we need more information? Right now we have 5 = (‘BC’ x height)/2.
Notice that–based on the given info–triangle BDE has 2/3 the base length of triangle ABC and is 1/2 as high. Thus, BDE has 1/3 the area of ABC:
- BDE = (1/3)ABC = 5/3
The second part is a little more tricky. Triangle ABD is the same height as ABC, but has only 2/3 the base length, so it’s area is 2/3 of ABC. This then implies triangle AEB is also 1/3 the area of ABC.
- AEB = (1/3)ABC = 5/3
Now draw a line between points D and F. Note the following:
- Triangle DFC is 1/3 the area of triangle BFC (the baseline DC is 1/3 the length of BC, and the same apex F).
- Triangle DEF and AEF have the same area (they have the same baseline lengths since AE=ED, and the same apex point F.
- Triangle BFC = BDE + DEF + DFC
Let X be the area of triangle AEF (= DEF). Then we can write BFC = 5/3 + x + (1/3)BFC, or BFC = (5/2) + (3/2)x.
The complete triangle ABC = BFC + AEB + AEF = (5/2) + (3/2)x + (5/3) + x = (25/6) + (5/2)x. But we were given that ABC = 5, so a little algebra gives x= 1/3. Thus, the area of triangle AEF is 1/3.
Wouldn’t triangle BDE only be half as high as triangle AED if >BDA was a right angle?
ETA: Just drew it out and realized the upright side varies directly with the length involved… nevermind. I need to crack out the Elements again.
My thought as well.
-FrL-
I don’t know what “upright side” means, but I just thought about it and realized that [spoiler]given a triangle ABC and a line segment perpendicular to AC with one endpoint at B and the other at a point D lying on AC, then the line parallel to AC and bisecting BD also bisects AB and BD.
This in turn implies that the line parallel to AC and bisecting BD also bisects any line segment from B to a point on AC. So if you have a line segment from one corner of a triangle to any point on the base, and bisect that line segment, then the distance of the point of bisection from the base is always half the hight of the triangle itself.[/spoiler]
-FrL-
Although “triangle AED” is meaningful and everything you said is reasonably interpretable as correct with it and all, I’m guessing you meant to say “triangle ABC” anyway.
“BD” at the end of my first spoilered paragraph should have said “BC.”
-FrL-
Yeah, that’s what I meant. I think ‘upright side’ might be terminology plucked from conics misapplied here. I meant your BD, of course.
This is already wrong. Nothing in the picture gives any indication of the height of either ABC or BDE. Now suppose we assume the trainagles are all right. This is not stated but it looks close at least. Even so, the triangles are evidently not similar and we are told that the ratio of the hypotenuses is 1:2 and of one of the legs of BDE is the same length as that of a certain segment emanating from A. But it is not half of any line in ABC.
Although I have not proved this, I suspect that the question is underdetermined. Unless there is a very clever diferent way of looking at it.
We are given that AE = ED, and that AED is a straight line. This makes the height of triangle BDE (the distance of point E from line BC) exactly half the height of ABC (the distance of A from the same line).
To elaborate, draw PQ parallel to BC such that it passes through E. If any line that runs from A to any point on BC is bisected by PQ, all such lines are (and we are told that AD, which is one such, is indeed so bisected). In particular, the height of ABC, being a perpendicular from BC to A, is bisected. Thus, although we know the height of neither, we do know that the height of BED is half the height of ABC.
To clear up my gloss, drop a perpendicular from E to the base BD. This forms a right triangle with hypoteneuse DE. Similarly, drop a perpendicular from A to the same base DE. This forms a second right triangle with hypoteneuse AD = 2*DE.
These two right triangles are similar (they are right triangles and share a common angle at point D). Thus, if AD is twice the length of BD, then the upright leg of the larger right triangle (i.e. the “height” of both ABD and ABC) is twice that of the smaller triangle (i.e. the “height” of BDE).
Just to help out, I figured I’d draw a slightly neater diagram here
CJJ* has a correct solution. I just came to point out a cute but not-much-taught idea for solving this sort of problem, which is the use of barycentric coordinates. Given a triangle ABC with area V, for a point X in the triangle at barycentric coordinates (a,b,c), normalized to a+b+c=1, the area of triangle BCX is |BCX|=aV, the area of triangle CAX is |CAX|=bV, and the area of triangle ABX is |ABX|=cV. If lines are drawn from each vertex through X to the opposite side (to K on BC, L on CA, and M on AB) then the areas of the subtriangles can also be found with barycentric coordinates; for example, the area of triangle BKX is a fraction c/(b+c) of the area of BCX, or |BKX|=(c/(b+c))aV = acV/(b+c). (Finding the areas of two of these subtriangles is the problem posed.)
The barycentric coordinates X=(a,b,c) satisfy the vector relation X=aA+bB+cC (where A, B, C are now interpreted as position vectors to the corresponding vertices; so X is the centroid of the system with point masses proportional to a, b, c at A, B, C. (Using this vector relation and the vector formulas for the area of a triangle, it is easy to verify that all of the above area relations hold.)
It is easy to find barycentric coordinates for the point E in the given problem. First, AE=ED implies that a=1/2 (A’s mass is enough to pull the centroid halfway between line BC and point A, so A has half the total mass). Now BD=2DC implies that c=2b (the centroid’s position off the line between itself and A is twice as far from B as from C, so C must have twice the mass of B), and since b+c=1-a=1/2 this means that b=1/6 and c=1/3, so E=(1/2,1/6,1/3). So now
|BDE| = acV/(b+c) = (5cm[sup]2[/sup])(1/2)(1/3)/(1/2+1/3) = 5/3 cm[sup]2[/sup]
and
|AEF| = bcV/(a+c) = (5cm[sup]2[/sup])(1/6)(1/3)/(1/2+1/3) = 1/3 cm[sup]2[/sup].