Help prove it, thank you, help prove it, thank you very much

In RT ABC, AB = AC +BC

D and E are the midpoints of BC and AC, AD=5, BE = 2 times the root number 10, AD=DC +AC=25, DC=BC 2

be²=bc²+ce²=40, ce=ac/2ac²=25-bc²/4

bc²=40-ac²/4

dc²+ac²+bc²+ce²=25+40bc²+ac²)5/4=65

bc²+ac²)=52=ab²

ab = 2 root number 13

Drawing. Let AE=x, cd=y, then AC=2X, BC=2Y in RT ADC, Pythagorean, (2x) 2+Y 2=25 in RT BCE, Pythagorean (2Y) 2+x 2=40, x 2+y 2=13

So, ab = root number 13

It may be a mistake, please do it again yourself, this way, the method is right)

1/2a）^2+b^2=(2∫10)^2

1/2b)^2+a^2=5^2

Let's do the rest for yourself, a b represents the length of ac bc.

The formula used is that the sum of squares of the two sides is equal to the third side.

Let cd=x, ce=y, and then use the Pythagorean theorem of the second degree.

x^2+(2y)^2=25

2x)^2+y^2=40

x^2=9y^2=16

Needless to say, the rest will be said.

From the inscription, it can be seen that in the process of increasing but less than flolow, the charge cannot move at a uniform speed and will always hit the two plates.

You don't want to say that p, p1, p2 are collinear when the modified vector p1p = ( +1) vector p1p2

op-op1= ( +1)(op2-op1)= ( +1)op2 - ( +1)op1op= ( +1)op2 - ( +1)op1 +op1=1 ( +1)op1 + ( +1)op2 (vector word omitted).

There must be other conditions, right? This alone cannot be calculated.

﹙a-b﹚(a+b﹚＝a²+ab-ab-b²=a²-b²

a+b)(a²－ab＋b²)＝a³－a²b＋ab²＋a²b－ab²﹢b³＝a³﹢b³

The focus of the problem is the distribution rate of multiplication, and it is recommended to look at the distribution rate of polynomials.

They all multiply and come out as soon as they are added.

a*b) t, which is considered as (a*b)*(a*b)*(a*b)*a*b) has t (a*b) multiplied, and when the parentheses are disassembled, it is a*b*a*b....a*b, there is group T Lazhou a*b; That is, there are t postures multiplied by a and t b multiplied; The bend is a t*b t.