How do you geometrically prove the content in the chapter on special triangles?

I don't know what you mean, but what do you have to prove?

rhs=secx-tanx cosecx-cotx =1 cosx-sin x cosx-cosx mega mill sinx = (sinx-sin x-cos x) sinxcosx =(cosecx-sinx-cot x) cotx (divide sin x on both denominator and nominator) =cosecx+cos x sinx+tanx =cosecx+cotx secx+tanx =lhs

Consider (csc x + cot x sec x + tan x) -sec x- tan x comic state csc x - cot x) =1 sin x + cos x sin x)(1 cos x) +sin x cos x - 1 cos x + sin x cos x)(1 sin x) +cos x sin x =1 sin x + 1 sin x + sin x cos x - 1 cos x + 1 cos x + cos x sin x =2 family finger bucket sin x + sin x cos x + cos x sin x =2 sin x + 1 (sin x)(cos x) is not equal to 0

Let 2sin = sin +cos, sin2 = sin cos solution: 4sin2 -2sin2 =1 from nanopulse 2-2, i.e., 2(1-cos2)-1-cos2)=1, 2cos2 =cos2

Cos4 -4cos4 = 2cos22 -1-4 (2cos22 -1).

2(2cos2α)2-8cos22α+3=3

<> this is my auxiliary line, here's how to prove it:

As shown in the figure, the corner bad=corner bad has been set'= Angular cad = Angular cad"= don't tell me you don't know why these two pairs of corners are equal), and the line segment ad=ad'=ad", hence the triangle ad'd"It is an isosceles triangle, hence the angle AD"d'=(180-2α-2β)/2=90-α-

Angle ad"d=(180-2) 2=90-, so angle dd"d'=angle ad"d - angle ad"d'= angular fad, i.e. angular dd" f=angular fad, hence a, f, d, d"Four-point contour type (decision theorem), and it is easy to know a, d, d", c four-point contour (this trembling mass guess does not need me to say), so there must be a, d, d", f, c five points contour, and notice that ac is for this or change the diameter of the circle, then there must be angle afc = 90 degrees. Finish.

Proof: The sum of the two sides of the triangle is greater than the third side.

Pa+PB AB, PB+PC BC, PA+PC AC After adding the above three formulas, it is obtained.

PA+PB+PB+PC+PA+PC AB+BC+AC is 2 (PA+PB+PC) Late book search AB+BC+AC, that is, PA+PB+PC 1 Posture 2 (AB+BC+AC), so that the title has been verified and the code has been revealed!