As shown in the figure, in the triangle ABC, D is a point on the edge AB, and ad = AC, De is parallel to BC, CD bisection angle EDF proves that AF bisection CD

As shown in the figure, in the triangle ABC, D is a point on the edge AB, and ad = AC, De is parallel to BC, CD bisection angle EDF proves that AF bisection CD

It is proved that because de ‖ BC, CD bisector EDF, then ∠ DCB = ∠ CDE = ∠ CDF, FD = FC
Ad = AC, so AF divides CD equally

It is known that ad is the middle line of triangle ABC, ab = AE, AC = AF, angle BAE = angle fac = 90 degrees

The ∠ AFD and ∠ AFE are equal
The proof is as follows:
from
AD=AB,
In other words, DAC = BAE
AC=AE,
(corner edge) provable Δ ADC ≌ Δ Abe
From this, we can deduce ∠ ADF = ∠ Abe, that is ∠ ADG = ∠ FBG (assuming that the intersection of AB and DF is point G)
And ∠ AGD = ∠ FGB (equal to vertex angle)
Then Δ AGD ∽ Δ FGB (two triangles are similar only if they have two equal angles)
Then ∠ BFG = ∠ DAG = 90 °,
Then segment be and segment DC are perpendicular to each other
That is to say, both BFD and CFE are right angles
This is the auxiliary line
Circle with the diameter of line BD and line CE respectively (the center of the circle is the midpoint of the two lines)
(please draw it yourself. I did, but it's not convenient to upload it.)
Then points a and F are on both circle BD and circle CE
(the diameter of a circle is a right angle to the circumference of the circle. Conversely, when the circumference of a circle is a right angle, the chord it faces is the diameter of the circle.)
If the inner chord ad of circle BD is opposite to ∠ AFD and ∠ abd, then ∠ AFD = ∠ abd = 45 degree
(in the same circle, the circular angles of the same chord are equal)
Similarly,
In the circle CE, the chord AE is opposite to ∠ AFE and ∠ ace, then ∠ AFE = ∠ ace = 45 degree
Therefore, AFD = AFE

Ad is the middle line of triangle ABC, AE is vertical AB, AF is vertical AC, AE equals AB, AF equals AC, and EF equals 2 ad.ef Vertical ad

Extend ad to g, make DG = Da, connect BG, easy to prove ≌ ABG ≌ AEF
∴AG=EF
∴EF=2AD

Ad is the midpoint of the triangle ABC, AE is perpendicular to AB, AF is perpendicular to AC, AE equals AB, AF equals AC

Link EF, take midpoint g of AB, link DG, take midpoint m of AE, midpoint n of AF, link Mn
D and G are both midpoint, so DG is parallel to AC
So angle AGD + angle GAC = 180
At the same time, angle GAC + angle EAF = 180
So (1) angle AGD = angle EAF
At the same time, (2) Gd = AC / 2 = AF / 2 = an
Similarly, (3) Ag = AB / 2 = AE / 2 = am
So the triangle AGD is congruent with the triangle Nam
So ad = Mn = EF / 2 (because Mn is the median line)
ef=2ad