It is known that in the isosceles triangle ABC, ab = AC, ad ⊥ BC at D, CG / / AB, BG intersect ad, AC in E and f respectively, and it is proved that be 2 = EF × eg

It is known that in the isosceles triangle ABC, ab = AC, ad ⊥ BC at D, CG / / AB, BG intersect ad, AC in E and f respectively, and it is proved that be 2 = EF × eg

It is proved that: connected CE. The Abe of the provable triangle is equal to the ace of the triangle, then the angle Abe = angle ace, be = CE. Because ab \ \ CG, the angle Abe = angle g, so the angle ace = angle g, and because the angle GEC is the common angle, so the triangle EFC is similar to the triangle ECG, so EC / EG = EF / EC, so EC ^ 2 = EF * eg, because EC = be, so be

In △ ABC, ad is the bisector of ∠ a, e is the midpoint of BC, EF / / AD is made through e, AB is intersected with G, and the extension line of Ca is at F. it is proved that BG = CF

First, it is ∠ a, which is an acute angle (otherwise, it intersects CA with F)
Use double length midline
Extend Fe to h, make ef = eh, connect BH
Prove that △ FCE is equal to △ HBE
CF = BH
From angle Bge = angle bhe, BG = BH
De BG = CF
Conclusion

It is known that in isosceles △ ABC, ab = AC, ad bisects ∠ BAC, intersects BC at point D, takes any point P (except point a) on line ad, passes through point P as EF ‖ AB, crosses AC, BC at point E and f respectively, makes PM ‖ AC, crosses ab at point m, and connects me (1) The results showed that the quadrilateral aepm was rhombic; (2) When point P is located, the area of rhombus aepm is half of that of quadrilateral efbm?

(1) It is proved that: ∵ EF ∥ AB, PM ∥ AC, ? AB = AC, ad bisection ∵ cab, ∵ CAD = ∵ bad, ∵ ad ⊥ BC (the property of three lines in one), ? bad = ∵ EPA, ∵ CAD = ∠ EPA, ∵ EA = EP, ? aepm is diamond

The triangle ABC, ab = AC, D is a point on the BC extension line, De is parallel to AC, the extension line of Ba is parallel to e, DF is parallel to the extension line of AB intersection AC, and the extension line of AC is at F, so AB = de-df

Because AB = AC, so angle B = angle ACB because De is parallel to AC, so angle EDC = angle ACB, so angle B = angle EDC, so EB = De, because De is parallel to AC and DF is parallel to AB, so the quadrilateral EDFA is a parallelogram, so AE = DF because ab = EB - AE (according to the above conclusion), so a

As shown in the figure, triangle ABC is an isosceles triangle, and point D is any point on the extension line of base BC. Crossing point D is de parallel AC, the extension line crossing Ba is at point E, DF is parallel to AB, and the extension line of intersection AB is at point F. what is the relationship between segment De, DF and segment AB? Why? Figure:

Because DF / / AE, de / / AC, afde is parallelogram, DF = AE, because ABC is isosceles triangle, EA = eg, ab = AC = Gd, that is, de-df = de-ea = de-ag = GD = ab

As shown in the figure in △ ABC, ab = AC, point D is on the extension line of Ba, point E is on AC, and the extension line of ad = AE, de intersects BC at point F, and DF ⊥ BC is proved

Proof: because ad = AE
So angle d = angle AED
Because angle AED = angle CEF
So angle d = angle CEF
Because AB = AC
So angle B = angle C
So triangle BFD is similar to triangle CFE (AA)
So angle BFD = angle CFE
Because the angle BFD + CFE = 180 degrees
So the angle BFD = angle CFE = 90 degrees
So DF vertical BC

As shown in the figure, in △ ABC, ab = AC, e is on AC, and the extension line of ad = AE, de intersects with BC at point F. verification: DF ⊥ BC

It is proved that as shown in the figure, am ⊥ BC is made by a to M,
∵AB=AC，
∴∠BAC=2∠BAM，
∵AD=AE，
∴∠D=∠AED，
∴∠BAC=∠D+∠AED=2∠D，
∴∠BAC=2∠BAM=2∠D，
∴∠BAM=∠D，
∴DF∥AM，
∵AM⊥BC，
∴DF⊥BC．

Given the triangle ABC, the vertex A is used as the vertical line of the bisector of angle B and angle c. AD is perpendicular to BD and D, AE is perpendicular to CE and e. it is proved that ED is parallel to BC, B and C

From the symmetry of angular bisector, it can be proved that △ abd ≌ △ MBD, so that G is the midpoint of AM; similarly, extending AE intersects BC at n, e is the midpoint of an, so De is the median line of △ amn,
You are not careful when you mark the points
Given triangle ABC, vertex A is used as angle B, bisector BD of angle C, vertical of CE, AG is perpendicular to BD and G, AH is perpendicular to CE and H. verify that GH is parallel to BC. you can reorganize it according to this description

In △ ABC, a (- 2,0) B (2,0), BC edge midline length ad = 3, find the trajectory equation of vertex C?

2|AD|^2+|BC|^2/2=|AC|^2+|AB|^2
18+[(x-2)+y^2]/2=(x+2)^2+y^2+16
x^2+12x+y^2=0
(x+6)^2+y^2=36(y≠0)
...
Let C (x, y)
D( (x+2)/2,y/2 )
|AD|=3
[(x+2)/2+2]^2+(y/2)^2=9
x^2+12x+y^2=0
(x+6)^2+y^2=36(y≠0)

Given that the edge ab of △ ABC is 4, if the length of the midline on BC edge is equal to 3, the trajectory equation of its vertex C is obtained

Establish a rectangular coordinate system, let a (0,0), B (4,0), C (x, y)
BC midpoint D ((x + 4) / 2, Y / 2)
(x+4)²/4+y²/4=3²
Trajectory equation: (x + 4) 2 + y 2 = 36, divided by (- 10,0), (2,0) two points