Given: F is the midpoint of AB. Prove: BD/CD = AE/EC

[Fouad1013]

Given: F is the midpoint of AB
Prove: BD/CD = AE/EC

I can't post the picture so I will describe it. It is a Triangle ABC with point E on Side AC and F on AB. There is a line drawn from F through E intersecting side BC extended at D.

I hope I have described the diagram sufficiently well. Please point out any missing details that you need.

Thank you!

 

[Denis]

Fouad, what grade are you in? That's sure not an easy one...

Let a = BC, b = AC, c = AB, d = CD, e = CE ; then AE = b - e

Draw EH parallel to BC, H on AB; let f = FH ; then BH = c/2 - f

Draw EG parallel to AB, G on BC; let g = GC ; then BG = a - g

So we have to prove that (a + d) / d = (b - e) / e

Line EH has created triangle AHE, which is similar to triangle ABC.
Line EG has created triangle EDG, which is similar to triangle FDB.

Step 1: isolate e using similar triangles ABC and AHE
c / b = (c/2 + f) / (b - e) ; simplify:
e = (bc - 2bf) / (2c)

Step 2: substitute e in (b - e) / e
(b - e) / e
= [b - (bc - 2bf) / (2c)] / [(bc - 2bf) / (2c)] ; simplify:
= (c + 2f) / (c - 2f)

Step 3: isolate g using (again) triangles ABC and AHE
c / a = (c/2 + f) / (a - g) ; simplify:
g = (ac - 2af) / (2c)

Step 4: isolate d using similar triangles FDB and EDG
(c/2) / (a + d) = (c/2 - f) / (d + g) ; simplify:
d = (ac - 2af - cg) / (2f)

Step 5: substitute g in above
d = [ac - 2af - c{(ac - 2af) / (2c) }] / (2f) ; simplify:
d = (ac - 2af) / (4f)

Step 6: substitute d in (a + d) / d
(a + d) / d
= [a + (ac - 2af) / (4f)] / [(ac - 2af) / (4f)] ; simplify:
= (c + 2f) / (c - 2f)

So:
see Step 2: (b - e) / e = (c + 2f) / (c - 2f)
see Step 6: (a + d) / d = (c + 2f) / (c - 2f)
So:
(b - e) / e = (a + d) / d

...yer lucky the Phoenix Suns were not playing tonight

I'm sure Soroban or TK or ..... can come up with something shorter/simpler :shock:

 

[Denis]

I worked with that some more; can be done much simpler:

Draw AG parallel to FD, G on extended BD;
then triangle ABG is similar to triangle FBD;
so, since AF = BF, then BD = DG.

Also, triangle ACG is similar to triangle ECD;
so DG/CD = AE/CE;
since DG = BD, then BD/CD = AE/CE.