As shown in the figure, in △ ABC, CF ⊥ AB is in F, be ⊥ AC is in E, M is the midpoint of BC, EF = 5, BC = 8, then the perimeter of △ EFM is () A. 13B. 18C. 15D. 21

As shown in the figure, in △ ABC, CF ⊥ AB is in F, be ⊥ AC is in E, M is the midpoint of BC, EF = 5, BC = 8, then the perimeter of △ EFM is () A. 13B. 18C. 15D. 21

∵ in △ ABC, CF ⊥ AB is in F, be ⊥ AC is in E, M is the midpoint of BC, BC = 8, ∵ MF = me = 12bc = 4, ∵ EF = 5, ∵ EFM circumference = 4 + 4 + 5 = 13, so a

As shown in the figure, in △ ABC, CF ⊥ AB is in F, be ⊥ AC is in E, and M is the midpoint of BC. (1) if EF = 4, BC = 10, find the perimeter of △ EFM;
Please prove it

(1) The center line on the hypotenuse of the triangle is equal to half of the hypotenuse, and EM = cm = BM is deduced. Similarly, FM = BM = cm is deduced. BM = cm = EM = FM = 5, so the perimeter of △ EFM is 14. From (1), BFM, CME and MEF are isosceles, ABC = 50 °, ACB = 60 ° and FMB = 80

As shown in the figure, in △ ABC, CF ⊥ AB and FBE ⊥ AC are at e, M is the midpoint of BC, EF = 4, BC = 6, then the perimeter of △ EFM is

∵CF⊥AB、BE⊥AC
∴RT△BCE、RT△BCF
∵ m is the midpoint of BC
Ψ me = MF = BC / 2 = 3 (right triangle midline characteristic)
The perimeter of EFM = me + MF + EF = 3 + 3 + 4 = 10

If O is the fixed point of the plane and P is in the plane determined by a, B and C and satisfies (OP OA) · (AB AC) = 0, then the trajectory of point P must pass △ ABC
A outer center B inner center C center of gravity D center of gravity
Why?

(op-oa)·(ab-ac)= ap.cb=0
So AP is perpendicular to CB
Answer: D
Have you ever studied vectors?
OP OA = AP, you know
Similarly, AB AC = CB
ap*cb=0
So the two vectors are perpendicular