D point on AB, ad: ab = 1:3, e point on BC, be: BC = 1:4, f point on Ca, CF: CA = 1:5, s △ def = 19, find s △ ABC?

D point on AB, ad: ab = 1:3, e point on BC, be: BC = 1:4, f point on Ca, CF: CA = 1:5, s △ def = 19, find s △ ABC?

forty-five point six

In equilateral △ ABC, D, e and F are the points on AB, BC and Ca respectively, ad = be = cf. ask the shape of △ def and explain the reason

Because ⊿ ABC is an equilateral triangle
So ∠ a = b = C, ab = AC = BC
And because ad = be = CF
So BD = CE = AF
From the congruent condition SAS of triangle
⊿ADF≌⊿BDE≌⊿CEF
So de = DF = EF
That is, ⊿ DEF is an equilateral triangle

In the triangle ABC, D is the midpoint of BC, e is the trisection of Ca, ad and be intersect at F, if the area of triangle ABC is

Fart! This problem is not complete

As shown in Figure 1, in equilateral △ ABC, points D, e and F are points on AB, BC and Ca respectively, and ad = be = CF
(2) : as shown in Figure 2, M is a point on line BC, connecting FM, making equilateral △ FMN on the right side of FM, connecting DM and en
(3) : as shown in Figure 3, change m as a point on the line BC to a point m as a point on the CB extension line, other conditions remain unchanged, and prove: DM = en
Please understand that the picture can't be passed on

ok
It is proved that: (1) ABC is an equilateral triangle,
In addition, ad = be = CF
∴△ADF≌△BED≌△CFE（SAS）,
∴DE=EF=DF,
The ∧ DFE is an equilateral triangle
(2) From (1), de = EF = DF,
In addition, MF = Mn = FM, ∠ DFM = ∠ EFM + 60 °, and ∠ EFN = ∠ EFM + 60 °,
∴∠DFM=∠EFN,
∴△DFM≌△EFN
∴DM=NE．
(3) Similarly, de = EF = DF, MF = Mn = FN,
In addition, MFD + MFE = 60 ° and MFE + EFN = 60 °,
∴∠MFD=∠EFN,
∴△MDF≌△NEF,
∴DM=EN．

In △ ABC and △ def, ab = De, BC = EF, AC = DF, a = D, B = e, C = f, there are several ways to judge the congruence of △ def?

There are three kinds of congruence: any two corners plus one side, side, two sides and their included angles
①C23×C13=9
② There's only one
③ Take the angle on both sides to determine C23 × 1 = 3
Total: 9 + 1 + 3 = 13

Are ab = De, AC / / DF, BC / / EF, △ ABC and △ def congruent?

Not necessarily congruent
When AB / / De, △ ABC and △ def are congruent

Ab ∥ De, AC ∥ DF, BC ∥ EF, prove ∥ def ∥ ABC
Pay attention to the similarity

Is there a graph? Different graphs may have different solutions

In △ ABC and △ def, ① AB = De, ② BC = EF, ③ AC = DF, and ④ ∠ a = ∠ d. There are three methods to determine △ ABC ≌ △ def from these four conditions______ Species

There are two methods to determine △ ABC ≌ △ def by SSS and △ ABC ≌ △ def by SAS

Given the triangle ABC, AC = 2, f is the midpoint of AC, angle a is 60 degrees, BF2 = ab × BC, find the length of ab

Let AC = x, BC = y
In △ ACF and △ ABC, the cosine theorem is applied respectively
CF^2＝X^2＋1－X
Y^2＝X^2＋4－2X
Because CF ^ 2 = AC * BC = XY
therefore
XY＝X^2＋1－X
Y^2＝X^2＋4－2X
Y = x + 1 / X-1 is obtained from xy = x ^ 2 + 1-x and substituted into another form to eliminate y
X^2＋2X－1＝0
Solution
X = √ 2-1 (negative root has been rounded off)
So AC = √ 2-1

As shown in the figure, △ ABC is an equilateral triangle, point D is any point on the edge of BC, de ⊥ AB is at point E, DF ⊥ AC is at point F. if BC = 2, then de + DF=______ ．

Let BD = x, then CD = 2-x. ∵ △ ABC is an equilateral triangle, ∵ B = ∠ C = 60 °. From the trigonometric function, ed = 32x, similarly, DF = 23 − 3x2. ∵ de + DF = 32x + 23 − 3x2 = 3