CDCG photography

Let's analyze it first. There are two right triangles (existing in triangle BCE) at the midpoint of this problem.

Generally speaking, the way to deal with the midpoint is to find another midpoint. As the saying goes, "One midpoint can't be beaten, but two midpoints are connected in a row."

You also need the length of DF, so it is easy to think of taking AE midpoint.

Then two classic right-angled triangles will think of using similar projective theorems and so on.

The following is the formal problem solving:

Solution: take AE midpoint g and connect DG.

Because AE = 2cm and G is the midpoint of AE, AG=GE= 1cm.

Because G and D are the midpoint of AE and AB respectively, DG∑BE and BE=2DG.

In the triangle CGD, EF/GD=CE/CG, because EF∨DG, CG=5cm, because CE = 4cm, EG = 1cm.

So EF/GD=4/5, that is, EF=(4/5)GD and EB=2GD, so EB/EF=5/2 let EB = 5k and EF = 2k(k >；; 0)

In RT triangle BCE, because ∠ ACB = 90, be ⊥ CD, there is ce 2 = ef * eb according to the projective theorem (triangle BCE∽ triangle CFE is used if you don't know the photography theorem).

Substitution data: 10k 2 = 16, the solution is k=(2√ 10)/5, so EB = 2 √ 10cm, EF = (4 √ 10)/5cm.

According to the projective theorem, if there is a solution of CF 2 = BF * EF, then CF=(4√ 15)/5cm is obtained.

In the triangle CGD, CE/EG=CF/FD is CF/FD=4 because EF∨DG, so DF = (1/4) CF = (√15)/5cm.

I hope my answer is helpful to you ~