This question popped up:
In triangle ABC, let D, E, and F be points on sides BC, AC, and AB such that BC=4CD, AC=5AE, and AB=6BF. If the area of triangle ABC is 120, find the area of triangle DEF.
Okay, since the Lone Ranger has gone on strike it is Geogebra to the rescue.
1) Use the polygon tool to draw triangle ABC.
2) Use the move tool to adjust the 3 vertices until poly1 = 120 or get as close as you can!
3)Use the circle tool to draw a cicle with C as the center and radius a/4.
4)Use the intersection tool and click side BC and the circle creating D
5)Hide the circle.
6)Use the circle tool to draw a cicle with A as the center and radius b/5.
7)Use the intersection tool and click side AC and the circle creating point E.
8) Hide the circle.
9)Use the circle tool to draw a cicle with B as the center and radius c/6.
10)Use the intersection tool and click side AB and the circle creating F.
11)Hide the circle.
12) You should have something like the first picture.
13)Use the polygon tool to draw triangle DEF. Should look something like the second picture.
14)Read in the variable pane what poly2 equals. If you have constructed a good first triangle ( one whose area is very close to 120 ) then you will get approximately 61 for an answer.
15)Experiment by dragging A,B and C around to different shaped traingles, always keeping the area of ABC = 120. When you are convinced that the smaller triangle is always close to 61 you are done!
You now have an approximate answer to a problem that has some people stumped.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
No great discovery was ever made without a bold guess.