In the triangle ABC, D is the midpoint of the BC side, and it is proved that AD & # 178; = 1 / 2 (AB & # 178; + AC & # 178;) - 1 / 4 (BC & # 178;) It's all vectors

In the triangle ABC, D is the midpoint of the BC side, and it is proved that AD & # 178; = 1 / 2 (AB & # 178; + AC & # 178;) - 1 / 4 (BC & # 178;) It's all vectors

In the triangle ABC, D is the midpoint of the BC side
AD²+CD²-AC²=2AD*CD*cos∠ADC
AD²+BD²-AB²=2AD*BD*cos∠ADB
∵∠ADC+∠ADB=180°,BD=CD=BC/2
∴AD²+BC²/4-AC²+AD²+BC²/4-AB²=0
∴AD²=1/2(AB²+AC²)-1/4（BC²）

As shown in the figure, in the triangle ABC, ab = AC, ﹥ 1 = ﹥ 2, prove: ab & ﹥ 178; = ad · AE

∵AB=AC
∴∠ABC=∠ACB
∵∠ABC=∠1+∠ABD
   ∠ACB=∠2+∠E
∵∠1=∠2
∴∠ABD=∠E
∵∠A=∠A
∴△ABD∽△AEB
∴AB/AE=AD/AB
∴AB²=AE ×AD

In △ ABC, ∠ cab = 90 °, ad ⊥ BC is D, 4AD = BC, find the degree of ∠ C

Take the midpoint e of BC and connect AE
Then AE = BC / 2 = 4AD / 2 = 2ad
In RT triangle ade, ∠ AED = 30 °
AE=CE
∠C=∠CAE
∠C+∠CAE=∠AED=30°
Therefore, C = 15 degree

As shown in the figure, in the triangle ABC, the angle cab = 90 degrees, ad is perpendicular to BC, AE is the midline on the side of BC, 4AD = BC, find the degree of angle C

∵SinB＝AD／AB
SinB＝Sin（90°－C）＝CosC
∴CosC＝AD／AB
　AD＝AB×CosC
∵SinC＝AB／BC＝AB／（4AD）＝1／（4CosC）
∴2SinC×CosC＝1／2
　Sin（2C）＝1／2
　2C＝30°
　∠C＝15°