Given that the triangle ABC, P is a point in the triangle, through P as PD vertical BC, PE vertical AC, PF vertical AB, the height of triangle ABC h, it is proved that PD + PE + pf = H

Connect AP, BP, CP
Let the length of an equilateral triangle be a and its area s
Then s = s (ABP) + s (BCP) + s (CAP)
=(1/2)×AB×PD+(1/2)×BC×PE+(1/2)×CA×PF
=(a/2)×PD+(a/2)×PE+(a/2)×PF
=(a/2)×(PD+PE+PF)
So PD + PE + pf = 2S / a 2S / a = H
So PD + PE + pf = H

In △ ABC, ab = 6, AC = 8, BC = 10, P is a moving point on edge BC, PE ⊥ AB is at e, PF ⊥ AC is at F, M is the midpoint of EF, then the minimum value of AM is______ .

∵ a quadrilateral afpe is a rectangle
∴AM=1
When 2AP, AP ⊥ BC, AP is the shortest, and am is also the shortest
When AP ⊥ BC, ⊥ ABP ∷ cab
∴AP：AC=AB：BC
∴AP：8=6：10
The shortest AP, AP = 4.8
When am is the shortest, am = AP △ 2 = 2.4

In △ ABC, ab = 6, AC = 8, BC = 10, P is a moving point on edge BC, PE ⊥ AB is at e, PF ⊥ AC is at F, M is the midpoint of EF, then the minimum value of AM is______ .

The triangle ABC is a right triangle with ab as the hypotenuse, AC = 4, BC = 3, P is the moving point on AB, and PE is perpendicular to AC to e, PF is perpendicular to BF to F, and the minimum value of EF is obtained

∵∠C=90°,PE⊥AC,PF⊥BC,
ν the quadrilateral PECF is a rectangle, ν EF = CP,
When p is on AB, the vertical segment is the shortest, ⊥ when PC ⊥ AB, PC is the shortest,
Then PC = 4 × 3 / 5 = 2.5
The minimum value of EF is 2.4

As shown in the figure, in the right triangle ABC, ∠ C = 90 °, AC = 1, BC = 2, P is the moving point on the hypotenuse ab. PE ⊥ BC, PF ⊥ Ca, then the minimum EF length of the line segment is______ .

Method 1: let EC = y, FC = X
∵∠C=90°，PE⊥BC，PF⊥CA，
The quadrilateral epfc is rectangular,
∴EP=FC=x；
∵AC=1，BC=2，
∴BE=2-y，
∵∠C=90°，PE⊥BC，
∴PE∥AC，
∴∠BPE=∠A，
And ∵ B = ∵ B,
∴2−y
2=x
1, i.e., y = 2 (1-x);
∵EF2=x2+y2
∴EF2=5（x-4
5）2+4
5（0＜x＜1），
When x = 4
At 5, EF is minimum=
Four
5=2
Five
5．
Method 2: connect PC,
∵PE⊥BC，PF⊥CA，
∴∠PEC=∠PFC=∠C=90°，
The quadrilateral ECFP is a rectangle,
∴EF=PC，
When the PC is the smallest, EF is the smallest,
That is, when CP ⊥ AB, the PC is the smallest,
∵AC=1，BC=2，
∴AB=
5，
The minimum value of PC is: AC · BC
AB=2
Five
5．
The minimum EF length of the line segment is 2
Five
5．

In the triangle ABC, the angle c = 90 degrees, the point P is on AB, PE vertical AC, PF vertical BC, the minimum EF length AB = 1, CF: BF = 2:1

Pf = AC / 3, EP = 2BC / 3, the square of EF = EP square + pf square = (2BC / 3) square + (AC / 3) square
Because BC square + AC square = AB square = 1, EF square = 1 / 9 + (BC Square) / 3 > 1 / 9
EF>1/3

As shown in the figure, in the right triangle ABC, ∠ C = 90 °, AC = 1, BC = 2, P is the moving point on the hypotenuse ab. PE ⊥ BC, PF ⊥ Ca, then the minimum EF length of the line segment is______ .

As shown in the figure, in RT △ ABC, ∠ C = 90 °, ABC = 30 ° and ab = 6. Point D is on the edge of AB, point E is a point on the edge of BC (not coincident with points B and C), and Da = De, then the value range of ad is______ .

Draw a circle with D as the center and the length of AD as the radius
① As shown in Figure 1, when the circle is tangent to BC, when de ⊥ BC,
∵∠ABC=30°，
∴DE=1
2BD，
∵AB=6，
∴AD=2；
② As shown in Figure 2, when the circle intersects with BC, if the intersection point is B or C, then ad = 1
2AB=3，
The value range of ad is 2 ≤ ad < 3

In RT △ ABC, ∠ C = 90 °, B = 35 ° and ab = 7, then the length of BC is () A. 7sin35° B. 7 cos350 C. 7cos35° D. 7tan35°

In RT △ ABC, CoSb = BC
AB，
∴BC=AB•cosB=7cos35°．
Therefore, C

In the RT triangle ABC, if the angle c = 90 ° and the angle B is 15 ° less than half of the outer angle of the angle c, what are the angles a and B equal to?

∵∠C=90°
The external angle of angle c is 90 degrees
∵ angle B is 15 ° less than half of the outer angle of angle C
∴∠B=90×0.5-15=30°
∵ triangle ABC is RT triangle
∴∠A=180°-90°-30°=60°

Given that the triangle ABC, P is a point in the triangle, through P as PD vertical BC, PE vertical AC, PF vertical AB, the height of triangle ABC h, it is proved that PD + PE + pf = H