看上去非常平常的几何线段和相等

isee

摘自《Which Way Did the Bicycle Go?》之几何部分（第 73 题）

如图 1 若 AB+BF=AD+DF，则 AC+CF=AE+EF.

战巡

1、由梅涅劳斯定理
\[\frac{DE}{AD}\cdot\frac{AC}{BC}\cdot\frac{BF}{EF}=1\]
\[\frac{BC}{AB}\cdot\frac{AE}{DE}\cdot\frac{DF}{CF}=1\]
相乘会得到
\[\frac{AC\cdot BF\cdot AE\cdot DF}{AD\cdot EF\cdot AB\cdot CF}=1\]

2、连$BD,CE$
既然$AB+BF=AD+DF$，就会有
\[AB-AD=DF-BF\]
两边平方即有
\[AB^2+AD^2-2AB\cdot AD=DF^2+BF^2-2DF\cdot BF\]
这其中按余弦定理，会有
\[AB^2+AD^2=BD^2+2\cos(\angle A)AB\cdot AD\]
\[DF^2+BF^2=BD^2+2\cos(\angle F)DF\cdot BF\]
代入上面就有
\[2AB\cdot AD(\cos(\angle A)-1)=2DF\cdot BF(\cos(\angle F)-1)\]
注意按（1）的结果，会有
\[\frac{AB\cdot AD}{DF\cdot BF}=\frac{AC\cdot AE}{EF\cdot CF}\]
这也就有
\[2AC\cdot AE(\cos(\angle A)-1)=2EF\cdot CF(\cos(\angle F)-1)\]
和上面类似，这里有
\[2AC\cdot AE\cos(\angle A)=AE^2+AC^2-EC^2\]
\[2EF\cdot CF\cos(\angle F)=EF^2+CF^2-EC^2\]
代入化简即有
\[(AC-AE)^2=(EF-CF)^2\]
这里可能产生两个结果，需要踢掉一个
懒得写了，自己弄吧...

Czhang271828

补一个原书做法 (摘自楼主的知乎主页)



作等腰三角 $\triangle BFB^\prime$, $\triangle CFC^\prime$,  $\triangle DFD^\prime$, $\triangle EFE^\prime$. 作相应的角平分线.

目标: 由题设 $AB^\prime =AD^\prime$ 证明结论 $AC^\prime$ 与 $AE^\prime$. 也就是证明 $B^\prime C^\prime =D^\prime E^\prime$.

注意到 $\angle FD^\prime B^\prime=\angle FC^\prime B^\prime$, 此时 $\{F,B^\prime ,D^\prime, C^\prime\}$ 共圆. 对称地, $\{F,B^\prime ,C^\prime, D^\prime, E^\prime\}$ 五点共圆.

由于 $\angle B^\prime FC^\prime=\angle D^\prime F^\prime E^\prime=\pi/2+\alpha-\beta-\gamma$ 是相等的圆周角, 从而 $B^\prime C^\prime =D^\prime E^\prime$.

isee

战巡 发表于 2024-9-6 03:31
1、由梅涅劳斯定理
\[\frac{DE}{AD}\cdot\frac{AC}{BC}\cdot\frac{BF}{EF}=1\]
\[\frac{BC}{AB}\cdot\frac{ ...

初步想过正弦定理，没想到利用余弦定理还比较麻烦~

hbghlyj

https://www.cut-the-knot.org/pythagoras/UrquhartsTheorem.shtml
Lemma

Let in ΔABC, angles at A, B, C measure 2α, 2β, and 2γ, respectively. Let p(ABC) be the perimeter of the triangle. Then

p(ABC) / BC = 2 / (1 - tanβ·tanγ).

Proof

We repeatedly use the Law of Sines and the addition formulae.

p(ABC) / BC = 1 + (AB + AC)/BC   = 1 + (sin2γ + sin2β) / sin 2α   = 1 + (sin2γ + sin2β) / sin(2β + 2γ)   = 1 + 2sin(γ + β)cos(γ - β) / 2sin(γ + β)cos(γ + β)   = 1 + (cosγ cosβ + sinγ sinβ) / (cosγ cosβ - sinγ sinβ)   = 2cosγ cosβ / (cosγ cosβ - sinγ sinβ)   = 2 / (1 - tanβ·tanγ).

Now let's return to Urquhart's theorem. Denote the angles as shown in the diagram.

Observe that in triangles AB'D and ACD side AD is common, so that, applying the Lemma to each of the triangles, we see that they have equal perimeters iff tanβ' tan(90° - γ') = tanγ tan(90° - β):

p(AB'D) = p(ACD)  iff tanβ' tan(90° - γ') = tanγ tan(90° - β)   iff tanβ' cotγ' = tanγ cotβ   iff tanβ' tanβ = tanγ tanγ'   iff p(ABD) = p(AC'D), as desired.

(An elementary synthetic solution appears elsewhere. Also, Michel Cabart noticed that the theorem extends to the non-degenerate conics in an absolutely simple manner, showing that even the most elementary theorem of geometry sprouts concealed relations to other pieces of mathematics.)