Four distinct points A, B, C, D on the plane satisfy:(vector AB-vector BC)*(vector AD-vector CD)=0, then △ABC shape? The four distinct points A, B, C, D the plane satisfy:(vector AB-vector BC)*(vector AD-vector CD)=0, then △ABC shape?

Four distinct points A, B, C, D on the plane satisfy:(vector AB-vector BC)*(vector AD-vector CD)=0, then △ABC shape? The four distinct points A, B, C, D the plane satisfy:(vector AB-vector BC)*(vector AD-vector CD)=0, then △ABC shape?

(Vector AB-vector BC)*(vector AD-vector CD)=(vector AB-vector BC)*(vector AD+vector DC)=(vector AB-vector BC)*vector AC=0
So vector AC is perpendicular to (vector AB-vector BC
Therefore, the diagonal line of the quadrilateral with AB and BC as adjacent sides is perpendicular to the diamond.
So △ABC is isosceles triangle

△ Vector AC=10, vector AD=5, vector AD=5/11 vector DB, CD*AB=0 in ABC Vector (AB-AC) Let ∠BAC=θ and cos (x)=4/5,-(π/2) △ Vector AC=10, vector AD=5, vector AD=5/11 vector DB, CD*AB=0 in ABC Find vector (AB-AC) Let ∠BAC=θ and cos (x)=4/5,-(π/2)

CD*AB =0 so CD is perpendicular to AB
Vector AD=5/11 vector DB, so point D is on AB edge
You draw the triangle ABC, the CD is perpendicular to AB and the point D is on the edge of AB.
Using the drawn triangles, AB =16 can be calculated from the Pythagorean theorem
Vector (AB-AC)=16-10=6

CD*AB =0 so CD is perpendicular to AB
Vector AD =5/11 Vector DB, so point D is on AB edge
You draw the triangle ABC, the CD is perpendicular to AB and the D point is on the AB side
Using the drawn triangles, AB =16 can be calculated from the Pythagorean theorem
Vector (AB-AC)=16-10=6

CD*AB =0 so CD is perpendicular to AB
Vector AD=5/11 vector DB, so point D is on AB edge
You draw the triangle ABC, the CD is perpendicular to AB and the D point is on the AB side
Using the drawn triangles, AB =16 can be calculated from the Pythagorean theorem
Vector (AB-AC)=16-10=6

There are four distinct points A, B, C, D on the plane satisfying (vector AB-vector BC).(vector AD-vector CD)=0, then triangle ABC is

(Vector AB - Vector BC).(Vector AD - Vector CD)=0
(-Vector BA-Vector BC).(Vector AD + Vector DC)=0
-(Vector BA + vector BC). Vector AC =0
Parallel quadrilateral ABCE with BA and BC as adjacent sides
Vector BA + vector BC = vector BD
Vector BE ●vector AC=0
BE AC
Diagonal lines of parallelogram ABCE are perpendicular to each other
ABCE is diamond
| AB|=|BC|
The triangle ABC is an isosceles triangle

D is the point on the plane of the triangle ABC, vector AD =1/2(vector AB + vector AC). If the module of vector BC =4, then what is the module of vector BD equal to Urgent

According to the condition that vector AD satisfies, D is the midpoint of BC. Since the module of vector BC is equal to 4, the module of vector BD is equal to 2.

According to the condition that the vector AD satisfies, D is the midpoint of BC.

If A and B are opposite to each other, then A+2A+3A+4A+5A+5B+5B+4B+3B+2B+B = Big brothers and, please help!

A+B=0
2A+2B=0
3A+3B=0
4A+4B=0
5A+5B=0
Last A+2A+3A+4A+5A+5B+4B+3B+2B+B=0

0A a+b 2a+b 2a+2b 3a+2b 3a+2b 3a+3b 4a+3b 4a+4b 5a+4b 5a+5b …… Find an expression. 0A a+b 2a+b 2a+2b 3a+2b 3a+2b 3a+3b 4a+3b 4a+4b 5a+4b 5a+5b …… Find an expression. According to the above rules, an expression requires only one variable, such as n

According to the column given by the Landlord, there are the following rules
F (n)= n/2*a+(n/2-1)*b when n is even
F (n)=(n-1)/2*(a+b) when n is odd

According to the column given by the main owner, there are the following rules
F (n)= n/2*a+(n/2-1)*b when n is even
F (n)=(n-1)/2*(a+b) when n is odd