If the vector Ba * (2 vector BC vector BA) = 0, then the shape of △ ABC is

If the vector Ba * (2 vector BC vector BA) = 0, then the shape of △ ABC is

BA.( 2BC-BA) =0
=> 2 BC.BA - |BA|^2 =0
|BA|^2 = 2 BC.BA
= 2|BC||BA| cos∠ABC
2|BC|cos∠ABC = |BA|
(isosceles triangle)

Vector AB + vector AC - vector BC + vector Ba need specific process, thank you,

The process omits two words of vector
BC=AC-AB,BA=-AB
Then: ab + ac-bc + Ba = AB + AC - (ac-ab) - AB
=2AB-AB=AB

In △ ABC, if the vector BA · (2 vector BC vector BA) = 0, then △ ABC must be
seek
A. Right triangle B, isosceles right triangle C, regular triangle D, isosceles triangle

D is C = 2acosb = acosb + bcosa, so acosb = bcosa, so a / b = cosa / CoSb has sine theorem, a / b = Sina / SINB, so cosa / CoSb = Sina / SINB, so sin (a-b) = 0, so a = B

Vector BA (2,3), vector Ca (4,7), vector BC (,)

Vector BC = vector Ba + vector AC = (- 2, - 3) + (- 4, - 7) = (- 6, - 10)

If Ba = (1,2), Ca (4,5), then BC=

BC=BA+AC=BA-CA=(-3,-3),

The graph is a pentagonal shape, ab = BC = CD = de = EA. It is known that a can walk 3 routes, B can walk 7 routes. If a and B start from point a at the same time, and walk clockwise, then a
The third time I caught up with B, it was on that side

Then a takes one part of the journey, B takes seven thirds of the journey, and B takes Four Thirds more than a,
Let AB = 1, then the perimeter of the Pentagon is 5,
3×5÷4/3=45/4,
That is, when B takes 15 more trips than a, a takes 45 / 4
That's two laps with 5 / 4,
So the third time B overtook a, it was on the side of BC

In the pentagonal ABCDE, ab = BC = CD = de = EA, and ∠ ABC = 2 ∠ DBE, it is proved that ∠ ABC = 60 degree

Because AE = AB, so ∠ Abe = ∠ AEB, the same as ∠ CBD = ∠ CDB, because ∠ ABC = 2 ∠ DBE, so ∠ Abe + ∠ CBD = ∠ DBE, because ∠ Abe = ∠ AEB, ∠ CBD = ∠ CDB, so ∠ AEB + ∠ CDB = ∠ DBE, so ∠ AED + ∠ CDE = 180 degrees, so AE parallel CD, because AE = CD, so quadrilateral AEDC is parallelogram, so D

As shown in Figure 3 is a regular pentagon, ab = BC = CD = de = EA
It is known that a can walk three times and B can walk seven times. If B starts from a at the same time and walks clockwise, which side will a catch up with B for the third time?

Where is the picture?

There is an equilateral triangle PQR in an equilateral pentagon ABCDE. QR coincides with AB, and PQR is flipped n times along its edges AB, BC, CD, De, EA in the Pentagon
If PQR returns to the original position at the same time after flipping n times, then n=——

n=15
to glance at

As shown in Figure 2, in the Pentagon ABCDE and Pentagon a1b1c1d1e1, if AB = A1B1, BC = b1c1, CD = c1d1, de = d1e1, EA = e1a1, please use as few conditions as possible to make them congruent. (write the conditions added, without explanation)

B = B1 and E = E1, the two groups of angles are equal