As shown in the figure, D and F are on the edge of AB of △ ABC, and ad = BF, de ‖ FG ‖ BC. De and FG intersect AC at points E and g respectively

As shown in the figure, D and F are on the edge of AB of △ ABC, and ad = BF, de ‖ FG ‖ BC. De and FG intersect AC at points E and g respectively

There is no picture, please give the picture

In the triangle ABC, BF is vertical to AC, CG is vertical to AB, F and G are perpendicular feet, D is the midpoint of BC, e is the midpoint of FG, and De is vertical to FG

In right triangle BCG
DG=BC/2
In the same way
DF=BC/2
So DG = DF
The middle line of the bottom of an isosceles triangle is the height
It's over

In triangle ABC, BF is perpendicular to AC, CG is perpendicular to ad, F and G are perpendicular feet, and D.E. is the midpoint of BC and FG respectively

It's CG vertical AB Let's make a picture first, and link DF and dgfd and Gd to be the center line of hypotenuse of right triangle BFC and BGC respectively. It can be seen that the length of the center line of hypotenuse of right triangle is equal to half of the length of hypotenuse, that is to say, DF and DG are equal to half of BC, so they are equal. In "equilateral" triangle DFG, De is the center line of FG at the bottom

In the triangle ABC, there is a point P, PE ⊥ AB, PF ⊥ AC. connecting Pb and PC, then ∠ ABP = ∠ ACP. M is the midpoint of BC, connecting me and MF

Prove: connect me, MF, BF, CE. Because PE is perpendicular to AB, PF is perpendicular to AC, so, angle BEP = angle CFP = 90 degrees, because angle ABP = angle ACP, so, angle BPE = angle CPF extends BP to Q, when AC is connected to Q, then, angle BPE = angle CPQ, so, angle CPF = angle CPQ, so, point F and point Q coincide, that is, BF and BP coincide, BF is the height on the edge of AC