It is known that: as shown in the figure, point D is a point on the edge AC of △ ABC, passing point D makes de ⊥ AB, DF ⊥ BC, e and F are perpendicular feet, then passing point D makes DG ∥ AB, intersecting BC at point G, and de = DF. (1) prove: DG = BG; (2) prove: BD bisects EF vertically

It is known that: as shown in the figure, point D is a point on the edge AC of △ ABC, passing point D makes de ⊥ AB, DF ⊥ BC, e and F are perpendicular feet, then passing point D makes DG ∥ AB, intersecting BC at point G, and de = DF. (1) prove: DG = BG; (2) prove: BD bisects EF vertically

The following: (1) connect BD \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\it's not easy

In the triangle ABC, ab = AC = 5, D is the point on BC, de / / AB intersects AC at e, DF / / AC intersects AB at F, and the perimeter of the quadrilateral FDEA is obtained

Because: De / / AB, DF / / AC, ab = AC
So: ∠ EDC = ∠ B = ∠ C = ∠ EDB
It is known that BF = DF, CE = De
So: AF + DF + de + AE = AF + be + CE + AE = AB + AC = 5 + 5 = 10
That is: the perimeter of the quadrilateral FDEA is 10

In the triangle ABC, ab = AC = a point d at any point on BC, do de parallel AB called AC and E, do DF parallel AC called AB and F, find the perimeter of quadrilateral AEDF

According to de / / AB, DF / / AC, AEDF is a parallelogram
So the perimeter of AEDF = AF + AE + de + DF
=2*(AF+FD)
AB = AC: angle B = angle c
DF / / AC: angle c = angle FDB
So angle B = angle FDB, so FB = FD
AEDF perimeter = 2 * (AF + FB) = 2 * AB = 2A
So the answer is 2A

As shown in the figure, △ ABC, ab = AC, ∠ BAC = 120 °, EF is the vertical bisector of AB, EF intersects BC with F, AB with E, and EF = 3, find the length of BF and CF

∵ AB = AC, ∵ BAC = 120 °, ∵ B = ∵ C = 12 (180 ° - 120 °) = 30 °, ∵ EF is the vertical bisector of AB, ∵ EF ⊥ AB, ∵ BF = 2ef = 2 × 3 = 6, connecting AF, ∵ EF is the vertical bisector of AB, ∵ AF = BF, ∵ BAF = ∵ B = 30 °, ∵ CAF = 120 ° - 30 ° = 90 °, ∵ CF = 2AF = 2 × 6 = 12

As shown in the figure, AE bisects ∠ BAC, BD = DC, de ⊥ BC, EM ⊥ AB, en ⊥ AC

It is proved that: connect be and EC, ∵ BD = DC, de ⊥ BC ∵ be = EC. ∵ AE bisection ≁ BAC, EM ⊥ AB, en ⊥ AC, EM = en, ≁ EMB = ≁ enc = 90 °. Be = EC, EM = enbe = ECEM = en, ≌ RT ≌ RT ≌ RT ≌ CNE (HL) ≁ BM = CN

It is known that in the RT triangle ABC, the angle c is equal to 90 degrees, and the vertical bisector of AB intersects AB at D and BC at E, AE.CD Intersection point O, prove ed bisector angle AEB, prove angle CDE is equal to angle CAE, prove OE / OC = AB / 2BC

Certification:
∵ De is the vertical bisector
‖ AE = be (the distance from the point on the vertical bisector to both sides is equal)
∴∠EAB=∠B
∵∠C=90°
The center line of the hypotenuse of a right triangle is equal to half of the hypotenuse
∴∠OCE=∠B
According to the sine theorem: OE / OC = sin ∠ OCE / sin ∠ OEC
∵∠OEC=∠EAB+∠B=2∠B
∴OE/OC=sin∠B/sin2∠B=sin∠B/(2sin∠Bcos∠B)=1/(2cos∠B)
∵cos∠B=BC/AB
∴OE/OC=AB/2BC
Can it solve your problem?

In the triangle ABC, point O is a moving point on the side of AC, and a straight line Mn / / BC is made through point O. suppose that the bisector ce of the intersection BCA of Mn is at e, and the bisector CF of the external angle BCA of Mn is at point F
When point O moves on edge AC, will the quadrilateral BCFE be a diamond? Proof
Thank you for everything

Let d be a point on the extension line of BC, according to the known, ∠ ECF = ∠ BCA / 2 + ∠ ACD / 2 = (∠ BCA + ∠ ACD) / 2 = 180 ° / 2 = 90 ° = RT ∠
If BCFE is a diamond, then FEC is an isosceles triangle, ∠ FEC = ∠ ECF = RT ∠
Meanwhile, the rhombus is a parallelogram, ∠ BEC = ∠ ECF = RT ∠
∠BEF=∠FEC+∠BEC=2*90°=180°
That is, B, e and F are three points and one line, which is contradictory to ef / / BC (that is Mn / / BC) and O being the last moving point of AC

As shown in the figure, in △ ABC & nbsp;, point O is a moving point on the edge of AC, and a straight line Mn ‖ BC is made through point O. suppose that the angular bisector of Mn intersecting ∠ BCA is at point E, and the outer angular bisector of intersecting ∠ BCA is at point F. (1) try to explain EO = fo; (2) when point O moves to where, is the quadrilateral aecf rectangular? (3) when the point O moves to where, and what condition does △ ABC satisfy, the quadrilateral aecf is a square? And explain the reason

(1) ∵ Mn ‖ BC, ∵ OEC = ∵ BCE, ∵ OFC = ∵ GCF, and ∵ CE bisecting ∵ BCO, CF bisecting ∵ GCO, ∵ OCE = ∵ BCE, ∵ OCF = ∵ GCF, ∵ OCE = ∵ OEC, ∵ OCF = ∵ OFC, ∵ EO = Co, fo = Co, ∵ EO = fo. (2) when point O moves to the midpoint of AC, quadrilateral aecf is rectangular (3) when the point O moves to the midpoint of AC, and △ ABC satisfies the condition that ∠ ACB is a right triangle, the quadrilateral aecf is a square. It is known from (2) that when the point O moves to the midpoint of AC, the quadrilateral aecf is a rectangle, and Mn ‖ BC is known. When ∠ ACB = 90 °, then ∠ AOF = ∠ COE = ∠ COF= The results show that: AOE = 90 °, AC ⊥ EF, aecf is square

As shown in the figure, △ ABC, point O is a moving point on the edge of AC, and a straight line Mn ‖ BC is made through point O. let the bisector of Mn intersection angle BCA be at point E and intersect the outer angle ACD of △ ABC

I haven't written all the questions. Please give me a supplementary

In △ ABC, point O is a moving point on the edge of AC, and a straight line Mn ‖ BC is made through point O. the bisector ce of Mn intersecting ∠ BCA is at point E, and the bisector CF of ACD intersecting ∠ BCA is at point F
Verification:
1) Verification: EO = fo
(2) When point O moves on AC, is the quadrilateral BCFE an l-diamond? If so, please prove your conclusion. If not, please explain the reason
(3) When the point O moves to where and the triangle ABC satisfies what conditions, the quadrilateral aecf is a square

(1) Because EC is the bisector of ∠ BCA, ∠ ECB = ∠ ace, so ∠ OEC = ∠ ace, so OE = OC. Similarly, OC = of, so EO = fo