It is known that points D, e and F are marked on each side AB, BC and Ca of triangle ABC such that AD / BD = be / CE = CF / AF = 1 / N. find the ratio of triangle def to triangle ABC (omitted, it's the ratio of their area) need to solve the problem process

It is known that points D, e and F are marked on each side AB, BC and Ca of triangle ABC such that AD / BD = be / CE = CF / AF = 1 / N. find the ratio of triangle def to triangle ABC (omitted, it's the ratio of their area) need to solve the problem process

S△ADF=1/2*AD*AF*sinA=1/2*1/(n+1)*AB*n/(n+1)*AC*sinA=n/(n+1)^2*S△ABC
Similarly, s △ BDE = s △ CEF = n / (n + 1) ^ 2 * s △ ABC
So s △ def = [1-3n / (n + 1) ^ 2] * s △ ABC = (n ^ 2-N + 1) / (n + 1) ^ 2

Let D, e and f be the points on BC, Ca and ab of the triangle, and DC equal to 2 times BD, CE equal to 2 times EA and AF equal to 2 times FB
Ask the relationship between "vector AD + vector be + vector CF" and vector BC,

Vector ad = AB + BD be = BC + CE CF = Ca + AF add AB BC CA = 0 CE and BF add up to 2 / 3 CB add BD (BD = 1 / 3bC) the final result is - 1 / 3 BC and BC reverse

Let D, e and f be the trilateral BC and Ca points of △ ABC respectively, and DC = 2bd, CE = 2EA and AF = 2fb, then the positions of AD + be + CF and BC?

Reverse parallel, select the basis vector AB, AC, or other AD.BE.CF . BC. Using vector operations

In the triangle ABC, the perpendicular of the bisector ad of the angle BAC passing through point C is D, de parallel to AB, and AC intersects at point e. it shows that AE is equal to CE?

It is proved that: ≠ DAF = ∠ DAC, ad = ad, ∠ ADF = ∠ ADC = 90 degree
∴△ADF≌△ADC
∫ DF = DC, and ∫ de ∥ AB,
A line passing through the midpoint of one side of the triangle and parallel to the other side bisects the third side

As shown in the figure, in △ ABC, ad bisects ∠ BAC, ad = AB, cm ⊥ ad intersects ad extension line at point M. verification: am = 12 (AB + AC)

It is proved that: lengthen am to N, make DM = Mn, connect CN, ∵ cm ⊥ ad, DM = Mn, ∵ CN = CD, ∵ CDN = DNC, ∵ DNC = DAB, ∵ ad = AB, ∵ B = ADB, ∵ B = ANC, ? bad = CAD, ? ADB = ACN, ∵ ANC = ACN, ∵ an = AC, ∵ AB + AC = AD + an = AD + am + M