Hurry, hurry! Help with the solution, junior high school math, proof questions

1. Because: RT ADE, RT ABC

So: a+ e= a+ b=90°, therefore: e= b...

1) And because :ed vertical ab, so: ade = fdb,。。

2) Synthesis (1) and (2) obtain: in RT Ade, FDB, A= DFB, So, ADE FDB.

2. Cd is the midline of the rt abc hypotenuse ab, cd=ad=db, ade fdb, ad ed=fd db (similar triangle, proportional to the corresponding edge), cd=ad=db, cd de=df cd, i.e.: cd 2=de*df.

The right triangle is fce, and the right triangle is ade, and the acute angle is common.

ADE FCE (Corner Angle).

Right triangle fce and right triangle fdb pair of vertex angles bfd = efc fdb fce (angle angle).

fdb∽ade

The midline of the hypotenuse of a right triangle is equal to half of the hypotenuse cd=ad=db fdb ade

ad/df=de/db

cd/df=de/cd

cd*cd=df*de

Upstairs wrong solution! Just the equality of the two sides is not enough to prove the congruence of two triangles!

By the sine theorem:

de sin dfe=df sin def and ac sin aec=ec sin eac, and df=ac, sin def sin aec, then de sin dfe ec sin eac, de=ec, sin dfe sin eac, so dfe eac, df ba, dfe bae, so bae eac, thus proves that ae is the angle bac bisector.

Proof: Since de ec, df ac, the triangle fde is equal to the triangle ace, so the angle dfe angle cae

Also dfiiba, get the angle efd angle eab, so the angle cae angle eab, so ae is the angle bac bisector.

Make parallel lines. Angle 1 = Angle 2

Because of parallelism. Angle 1 = Angle 6

Angle 2 = Angle 6fa = Fe

Because the midpoint and parallel.

Easy to get fa=fb

fe = fb angle 3 = angle 5

This is the bisector.

Passing e as a parallel line of AD, intersecting AB at the point F Because E is the midpoint, F is also the midpoint of AB, i.e., af=bf

Angle dae = angle fae At the same time the angle dae = angle aef (the wrong angle is equal) So, the angle fae = the angle aef....The triangle AFE is an isosceles triangle, so AF=EF

And because af=bf

So: ef=bf

So the triangular BFE is an isosceles triangle.

So the angle fbe = the angle feb

And because the angle feb = angle ebc (the inner error angle of the parallel lines is equal), the angle fbe = the angle ebc

So it's the angular bisector.

Over e as EF ad bc, f on ab. Because E is the midpoint of the line segment DC, F is the midpoint of AB. Because ae is the bisector of bad, so the angle eda=angle eaf, and ef ad, so fea= ead= eaf, so fe=fa, and f is the midpoint of ab, so fe=fb, so feb= fbe, and ef bc, so feb= ebc, so ebc= fbe, i.e. be is the bisector of abc.