In △ ABC, ad bisects ∠ BAC ∠ C ＞ B to prove 1) ∠ ADC = 90 ° - 1 / 2 (∠ C - ∠ b) 2) ∠ ADC = 1 / (∠ ACE + ∠ b)

In △ ABC, ad bisects ∠ BAC ∠ C ＞ B to prove 1) ∠ ADC = 90 ° - 1 / 2 (∠ C - ∠ b) 2) ∠ ADC = 1 / (∠ ACE + ∠ b)

∠ADC=1/2∠A+∠B
∠ADC=180°-1/2∠A-∠C
2∠ADC=180+∠B-∠C
So ∠ ADC = 90 ° - 1 / 2 (∠ C - ∠ b)

It is known that e.f.g.h is respectively the midpoint of AB, BC, CD and Da of the space quadrilateral ABCD. The four points of E, F, G and H are proved by vector method

Link AC
Vector eg = eh + Hg
According to the median line, the vector Hg = 1 / 2 AC and the vector EF = 1 / 2 AC can be obtained
The vector EF = Hg
Vector eg = eh + ef
Four points coplanar

In the parallelogram ABCD, the module of AB is 4, the module of ad is 3dab is 60 degrees. Find the vector ad times the vector BC, the vector AB times the vector CD, the vector AB times the vector da
Such as the title

Don't tell me that you can't do it even if it's so simple. You can draw a picture, judge the angle between vectors, and use the loud multiplication formula to calculate

In parallelogram ABCD, vector AB + vector AC + vector Da + vector CD + vector BC + vector BD=
1, vector Ba 2,2, vector DC 3. Vector ad 4,2, vector BC

AB + AC +DA + CD +BC +BD
=(AB +BC) +(AC+CD) + (BD+DA)
= AC +AD+BA
= AC +AD + CD ( BA = CD)
= AD +AD
=2AD
=2BC (AD=BC)
ans:(4)