As shown in the figure, in RT △ ABC, ∠ BAC = 90 °, ab = AC, and points m and N are on edge BC. (1) as shown in Figure 1, if am = an, prove: BM = CN; (2) as shown in Figure 2, if M and n are any two points on edge BC, and satisfy ∠ man = 45 °, then are line segments BM, Mn and NC possible to make equation Mn2 = BM2 + NC2 stand? If yes, please prove; if not, please give reasons

As shown in the figure, in RT △ ABC, ∠ BAC = 90 °, ab = AC, and points m and N are on edge BC. (1) as shown in Figure 1, if am = an, prove: BM = CN; (2) as shown in Figure 2, if M and n are any two points on edge BC, and satisfy ∠ man = 45 °, then are line segments BM, Mn and NC possible to make equation Mn2 = BM2 + NC2 stand? If yes, please prove; if not, please give reasons

(1) It is proved that: ∵ AB = AC, ∵ B = ∵ C. ∵ am = an, ∵ amn = ≌ anm. That is to say, ∵ AMB = ≌ ANC. (1 point) in △ ABM and △ can, ∵ AMB = ≌ ANC ≌ B = cab = AC ≌ ABM ≌ can (AAS). (2 points) ≌ BM = CN. (1 point) in addition, it is proved that through point a is ad ⊥ BC, and perpendicular foot is point D

m> It is proved that a / (a + m) + B / (B + m) is greater than C / (c + m)

Method 1
a. B, C, and m are positive numbers
So (a + m) (B + m) (c + m) are all greater than 0
To prove a / (a + m) + B / (B + m) > C / (c + m)
That is, a (B + m) * (c + m) + B (a + m) * (c + m) > C (a + m) (B + m)
ABC + ABM + ACM + AMM + ABC + ABM + BCM + BMM ABC ACM BCM CMM > 0
That is ABM + AMM + ABC + ABM + BMM CMM > 0
And because a + b > C mm > 0
So AMM + BMM > CMM
So ABM + AMM + ABC + ABM + BMM CMM > 0
Get proof
Method 2
A / (a + m) + B / (B + m) - C / (c + m)
=[a(b+m)(c+m)+b(a+m)(c+m)-c(a+m)(b+m)]/[(a+m)(b+m)(c+m)]
Because the three sides of triangle ABC are a, B, C > 0, and M is a positive number
So the denominator [(a + m) (B + m) (c + m)] > 0
And because a (B + m) (c + m) + B (a + m) (c + m) - C (a + m) (B + m)
=abc+abm+acm+am^2+abc+bam+bcm+bm^2-abc-cam-cbm-cm^2
=abc+(abm+bam)+(am^2+bm^2-cm^2)
Because a + b > C (the sum of the two sides of the triangle is greater than the third side)
So am ^ 2 + BM ^ 2 = (a + b) m ^ 2 > cm ^ 2
So (am ^ 2 + BM ^ 2-cm ^ 2) > 0
abc+(abm+bam)>0
So a / (a + m) + B / (B + m) > C / (c + m)

It is known that the three sides of △ ABC are the reciprocal of a, B and C, and the reciprocal of a, B and C is an arithmetic sequence. It is proved by analysis that ∠ B is an acute angle

Analysis method: to prove that the "B" is an acute angle, that is, to prove CoSb > 0, that is, to prove (a (A & #35;178; + C & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\﹥ 178; > 4A & ﹥ 178; C & ﹥ 178;, considering a & ﹥ 178; + C & ﹥ 178; ≥ 2Ac, (a + C) &﹥ 178; ≥ 4ac
So (A & # 178; + C & # 178;) (a + C) &# 178; ≥ 8A & # 178; C & # 178; > 4A & # 178; C & # 178;), so ∠ B is an acute angle
As a result of the comprehensive method, the general method is as follows: 2 / b = 1 / A / A + 1 / C, (A / b = 1 / A, a / b = 1 / A / A / C, \35;178; + C \#35;178; (; (C, \###35;\35;\35;\\\\\35;\178; (; (; (C, \\\\\\\\\\\\\\\\\\\\\\\\2 / b = 1 / A + 1 / C ﹥ A & ﹥ 178; + C & ﹥ 178; > b & ﹥ 178;, i.e. CoSb > 0, ﹥ B is the acute angle
In fact, the comprehensive method is to write the analytic method in reverse order. The comprehensive method is simple and easy to use. People often use the analytical method to analyze and use the comprehensive method to write the proof process

The three sides of the triangle ABC are a, B, C respectively. The middle line on the side BC is marked as m, which is used to prove the string theorem. M = 1 / 2 radical 2 (b ^ 2 + C ^ 2) - A ^ 2
Hurry, please, hee hee

According to the cosine theorem, cos B = (a ^ 2 + C ^ 2-B ^ 2) / 2Ac,
m^2=AD^2=c^2+(a/2)^2-2*c*a/2*cos B,
Replace the first formula with the second one and simplify it
The length of bisector and high line of triangle angle can be obtained by the same method