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As shown in the figure: in isosceles RT △ ABC, ∠ C = 90 degrees, AC = 8, f is the midpoint of AB edge, points D and e move on AC and BC sides respectively, and keep ad = CE, connect We need the whole proof process. Thank you
As shown in the figure: in isosceles RT △ ABC, ∠ C = 90 degrees, AC = 8, f is the midpoint of AB side, points D and e move on AC and BC sides respectively, and keep ad = CE, connect De, DF, EF. In the process of motion change, the following conclusions are obtained: 1, △ DFE is isosceles right triangle. 2. Quadrilateral cdef cannot be square. 3. The length of De is the smallest
As shown in the figure RT △ ABC, ∠ C = 90 ° a = 30 ° points D, e are on AB, AC and de ⊥ AB respectively
Using Pythagorean theorem and similarity of triangle
A kind of
Let de = 1, then AE = 2
According to the area ratio equal to the square of similarity ratio, de: BC = 1: √ 2, BC = √ 2
So AC = √ 6
So Ce: AE = (√ 6-2): 2
As shown in the figure, in RT △ ABC, ∠ C = 90 °, ab = 50, AC = 30, D, e, f are the midpoint of AC, AB and BC respectively. Point P starts from point D and moves along the broken line de-ef-fc-cd at a speed of 7 units per second; point Q starts from point B at a speed of 4 units per second in the direction of Ba, and passes through point Q as the ray QK ⊥ AB, and the intersecting broken line bc-ca starts from point g. points P and Q at the same time, when point P makes a circle When point d stops moving, point Q also stops. Set point P and Q to move for T seconds (T > 0) (1) The distance between D and F is______ ; (2) Can X-ray QK divide a quadrilateral cdef into two parts of equal area? If it can, find the value of T; if not, explain the reason; (3) When the point P moves to the broken line ef-fc, and the point P falls on the ray QK, calculate the value of T; (4) Connect PG, when PG ∥ AB, write the value of t directly
(1) In RT △ ABC, ∠ C = 90 ° AB = 50,
∵ D, f is the midpoint of AC, BC,
/ / DF is the median line of △ ABC,
∴DF=1
2AB=25
So the answer is: 25
(2) Yes
As shown in Fig. 1, connect DF and make FH ⊥ AB at point h through point F,
∵ D, f is the midpoint of AC, BC,
∧ de ∥ BC, EF ∥ AC, quadrilateral cdef is rectangular,
When QK passes through the midpoint o of DF, QK divides the rectangular cdef into two parts with equal area
In this case, QH = of = 12.5. From BF = 20, △ HbF ∽ CBA, Hb = 16
So t=QH+HB
4=12.5+16
4＝71
8．
(3) (1) when point P is on EF (26
7 ≤ t ≤ 5),
As shown in Figure 2, QB = 4T, de + EP = 7T,
From △ PQE to △ BCA, 7T − 20 is obtained
50＝25−4t
30．
∴t=421
41；
② When point P is on FC (5 ≤ t ≤ 76
7) When,
As shown in Fig. 3, QB = 4T, thus Pb = QB
cos∠B=4t
Four
5=5t，
From pf = 7t-35, BF = 20, 5T = 7t-35 + 20
T = 71
2；
(4) As shown in Figure 4, t = 12
3. As shown in Figure 5, t = 739
43．
(Note: PG ∥ AB can be divided into the following situations: when 0 ﹤ t ≤ 26
At 7:00, point P goes down and point G goes up. It can be seen that there is a time when PG ∥ AB exists,
As shown in Fig. 4, when point G continues to go up to point F, t = 4, while point P goes down to point E and then goes up along EF. It is found that there is no PG ∥ AB when point P moves on EF; 5 ≤ t ≤ 76
At 7, points P and G are on FC, and there is no PG ∥ AB; because point P reaches point C earlier than point G and continues to go down along CD, so the
Seventy-six
When PG ∥ AB exists in 7 ﹤ t ﹤ 8, as shown in Fig. 5, when 8 ≤ t ≤ 10, the points P and G are on CD, and there is no PG ∥ AB)
In RT △ ABC, ∠ C = 90 °, AC = 3, BC = 4, then AB =, the high Cd on the hypotenuse=
According to the Pythagorean theorem,
AB² = AB² + AC² = 3² + 4² = 25 = 5²
So AB = 5
Because s △ ABC = (1 / 2) × AC × BC = (1 / 2) × ab × CD
So (1 / 2) × 3 × 4 = (1 / 2) × 5 × CD
CD = 12/5
It is known that CD is the height on the hypotenuse ab of RT △ ABC. The lengths of AC, BC and ab are B, a and C respectively, and CD = H ⒈c+h〉a+b. Three sides of a+b, c+h and h can form right triangle
prove:
(1)
∵(a+b)²=a²+2ab+c²,
∵ a ∵ a ∵ a ∵ B ∵ B ∵ a ∵ B ∵ C ᙽ 2Ab = 2CH (from area)
∴(c+h)²＞(a+b)²
∴c+h＞a+b
(2)
∵(a+b)²=a²+2ab+c²,(c+h)²=c²+2ch+h²
∴(c+h)²=(a+b)²+h²
A right triangle can be formed by taking a + B, C + H and h as three sides
As shown in the figure, in RT △ ABC, ∠ C = 90 ° CD ⊥ AB in D, AC = 3, BC = 4, ab = 5, find the length of CD, and find s △ ABC
If C = 90 ° CD ⊥ AB is in D, then ∽ ACD
AC/CD=AB/AC→3/CD=5/3→CD=9/5
S△ABC=AC*BC=3*4=12
In RT △ ABC, if the angle c = 90 ° AB = 10, the length ratio of BC to AC is 3:4, then BC = ---- AC=------
The length ratio of BC to AC is 3:4, BC = 3x, AC = 4x: AC 2 + BC 2 = AB 2, (4x) 2 + (3x) 2 = 10? 2, 16x? 2 + 9x? 2 = 100 / 25 = 4, x = 2, (x = - 2 rounding), BC = 3x = 3 * 2 = 6; AC = 4x = 4 * 2 = 8;
In RT △ ABC, ∠ C = 90 °, BC: AC = 3:4, ab = 10. Find the length of AC and BC
Let BC = 3x, then AC = 4x, according to Pythagorean theorem
AC^2+BC^2=AB^2
therefore
25x^2=100
X = 2 (- 2 does not conform to the question and will be omitted)
So, AC = 6
In the RT triangle ABC, ∠ C = 90 °, BC: AC = 3:4, ab = 10, find the length of AC and ab
Let the side length of BC be 3x and AC be 4x. According to Pythagorean theorem, we can get (3x) ^ 2 + (4x) ^ 2 = 10 ^ 2, then we can solve X
In RT Δ ABC, ∠ C = 90 °, AC = 15, ab-bc = 9
Let BC = x, then AB = x + 9
From Pythagorean theorem: AB 2 = AC 2 + BC 2
（9+x)²=15²+x²
X=8
So BC = 8, ab = 17
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