听的是B站UP主唐小谦的解析几何课,万年的老计算机专业人也想学数学OWO。
1. 课程介绍与要求
前面都是老师的课程介绍,从板书证明开始记
【证明】在三角形ABC中,D为BC边的中点,证明: ∠ ABC = 9 0 ∘ \angle \text{ABC}=90^{\circ} ∠ABC=90∘的充要条件是 AD = 1 2 BC \text{AD}=\frac{1}{2}\text{BC} AD=21BC.
【证】先证充分性,
因为D为BC边的中点,所以BD=DC,
因为 AD = 1 2 BC \text{AD}=\frac{1}{2}\text{BC} AD=21BC,所以 AD = BD = DC \text{AD}=\text{BD}=\text{DC} AD=BD=DC
所以三角形ADB和三角形ADC是等腰三角形
所以 ∠ DBA = ∠ DAB \angle \text{DBA}=\angle \text{DAB} ∠DBA=∠DAB, ∠ DCA = ∠ DAC \angle \text{DCA}=\angle \text{DAC} ∠DCA=∠DAC
故 ∠ BAC = ∠ DAB + ∠ DAC = ∠ DBA + ∠ DCA \angle \text{BAC}=\angle \text{DAB}+\angle \text{DAC}=\angle \text{DBA}+\angle \text{DCA} ∠BAC=∠DAB+∠DAC=∠DBA+∠DCA
由于 ∠ BAC + ∠ DBA + ∠ DCA = 2 ( ∠ DBA + ∠ DCA ) = 18 0 ∘ \angle \text{BAC}+\angle \text{DBA}+\angle \text{DCA}=2(\angle \text{DBA}+\angle \text{DCA})=180^{\circ} ∠BAC+∠DBA+∠DCA=2(∠DBA+∠DCA)=180∘
所以 ∠ DBA + ∠ DCA = ∠ BAC = 9 0 ∘ \angle \text{DBA}+\angle \text{DCA}=\angle \text{BAC}=90^{\circ} ∠DBA+∠DCA=∠BAC=90∘
再证必要性(已知: ∠ BAC = 9 0 ∘ \angle \text{BAC}=90^{\circ} ∠BAC=90∘)
以D点为原点,以向量 B C ⃗ \vec{BC} BC 方向为 x x x轴正方向,以垂直于BC边向上的方向为 y y y轴正方向建立如下坐标系:
则三角形中各点坐标分别为 A ( x , y ) , B ( − R , 0 ) , D ( 0 , 0 ) , C ( R , 0 ) \text{A}(x,y),\text{B}(-R,0),\text{D}(0,0),\text{C}(R,0) A(x,y),B(−R,0),D(0,0),C(R,0)
由于 ∠ BAC = 9 0 ∘ \angle \text{BAC}=90^{\circ} ∠BAC=90∘,且 A B ⃗ = ( x + R , y ) , A C ⃗ = ( x − R , y ) \vec{AB}=(x+R,y),\vec{AC}=(x-R,y) AB =(x+R,y),AC =(x−R,y)
则 A B ⃗ ⋅ A C ⃗ = 0 \vec{AB}\cdot \vec{AC}=0 AB ⋅AC =0,即 ( x + R ) ( x − R ) + y 2 = 0 (x+R)(x-R)+y^{2}=0 (x+R)(x−R)+y2=0
亦即 x 2 + y 2 = R 2 x^{2}+y^{2}=R^{2} x2+y2=R2
说明点A在以点D为圆心,半径为 R R R的圆上
则 AD = BD = BC = R \text{AD}=\text{BD}=\text{BC}=R AD=BD=BC=R
又因为 BC = 2 R \text{BC}=2R BC=2R
则 AD = 1 2 BC \text{AD}=\frac{1}{2}\text{BC} AD=21BC