First of all, it is not easy to prove the construction method of equilateral triangle.
I noticed that the framework of graphics was fixed, so I turned to algebra. Algebra is actually a routine. I hope it will help LZ ~
Let's start with some obvious conclusions:
CA=CB
Application of Sine Theorem in δδCDB;
CD/sin20 =CB/sin40 ……①
Using sine theorem in δδCEA;
CE/sin 10 =CA/sin30 ……②
①/② De
CD/CE
=sin30sin20/(sin40sin 10) (the following angles are omitted)
= 1/2 * 2 sin 10 cos 10/(sin 40 sin 10)
=cos 10/sin40
Application of sine theorem in δδCDE;
CD/CE=sin(30+x)/sin(x+ 10)
Relative availability:
sin(30+x)/sin(x+ 10)= cos 10/sin 40……③
Okay, solve the equation. After this problem is finished, here are the routines:
Step one:
Guess the answer ... an embarrassing and effective way, just make a standard chart and test it, 20.
The test paper says that X = 20 is brought in.
3 equivalent to
sin50sin40=sin30cos 10
be equal to
1/2[cos 10-cos 90]= 1/2 * cos 10
Equivalent to1/2cos10 =1/2cos10.
The above formula obviously holds, so 20 is the root of the equation.
Step 2: Monotonicity.
Note that the right side of ③ is a constant. As long as it is proved that the left side is strictly monotonous in the definition domain, there is only one solution.
Let the midpoint be o, obviously OE & gtOD,
∠ doe = 50, complementary angle is140, and x should be less than half of 70.
The domain is (0,70) (maybe it can be reduced, but it is enough).
sin(30+x)/sin(x+ 10)
=[sin(x+ 10)cos 20+sin 20 cos(x+ 10)]/sin(x+ 10)
=cos20+sin20/tan(x+ 10)
Strictly monotonically decreasing in the definition domain.
Here, the certificate will be done!
This geometric method is not easy to do, and it is almost invincible with this routine!