Find a math proof question for junior high school and a math proof question for junior high school

Crossing the point A is the angular bisector of A, intersecting BC at the point D, and then passing the point D is the perpendicular line of AB, and the perpendicular foot is E, then BAD=1 2 BAC

From the known b=1 2 bac, we get b= bad, so we can know that bad is an isosceles triangle and ab is the bottom edge.

De ab again, we can know that e is the midpoint of ab, that is, ae=1 2ab=ac and bad= cad, we get aed acd, so c= aed=90°

Connect AD

Because fad= deb af=be ad=db, afd is equal to deb

So fda= edb

Because edb+ ade=90°

So fde=90°

Proof is that the extension of CE is at AB and F, AD is the bisector of the angle, and it is the height of the triangle ACF, so the triangle AFC is an isosceles triangle, and AF= is the midpoint of CF. Because M is the midpoint of BC, ME is the median line of the triangle BCF, and ME BF is ME AB

me=1/2bf=1/2(ab-af)=1/2(ab-ac)

Oh, my ladygaga this is the third year of junior high school, right?

Solution (1) The bisector of b and c be, cf intersects at the point i

IBC+ ICB=1 2 (ABC+ ACB) and BIC=180°- (IBC+ ICB) BIC=180°-1/2 (ABC+ ACB) holds.

2) abc+ acb=180°- a1 2 (abc+ acb)=90°-1 2 a, bic=180°-one-half (abc+ acb) can be obtained from the conclusion of (1).

180°-（90°-1/2 ∠a）

i.e. bic=90°+one-half a

The process of solving the problem is so decent, so you're welcome, just let me do it thanks!

Do the extension of the FG AC cross AD at G

In a right-triangle ADC, E is the midpoint of AC.

then de=ae

ade=∠dae

then in the triangle fdg.

fgd=∠dae=∠ade=∠fdg

then fg=df

and C+DAC=90° GAF+ DAC=90° then C= GAF

then ab:ac=fg:af

fg=df

then ab:ac=df:af