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As shown in the figure, in △ ABC, ∠ BAC = 90 °, extend Ba to point d so that ad = half AB, points E and F are the midpoint of edge BC and AC respectively. Find DF = AE
Connect ef
∵ E and F are the midpoint of side BC and AC respectively
The EF is the median line of △ ABC
∴EF=1/2AB
EF∥AB
∵AD=1/2AB
∴AD=EF
∵EF∥AD(AB)
The ADFE is a parallelogram
∴DF=AE
In the triangle ABC, ∠ BAC = 90 ° and extend Ba to point d so that ad is equal to half of AB, points g, e, f are the midpoint of AB, BC and AC respectively, and DF = be is proved
In this paper, we prove that the triangle ADF is all equal to GBE, the theorem of edge, corner and edge, ad = GB, angle DAF = angle Bge is equal to 90 degrees, AF = Ge, so DF = be
As shown in the figure, triangle ABC is a right triangle with angle BAC = 90 degrees. D is the midpoint of hypotenuse BC, e and F are points on AB and AC respectively, and De is perpendicular to DF AB is not equal to ac. it is proved that be square + CF square = EF square
Extend ed to G so that DG = De, connect FG, BG because BD = DC, ed = DG, angle BDE = CDG, so triangle BDE and CDG are congruent, so be = CG, angle EBD = GCD, because ed = DG, FD is vertical eg, EF = FG because angle a = 90 degrees, so angle B + ACB = 90 degrees, so angle GCD + ACB = 90 degrees, so angle GCF = 90 degrees, so FG square = CG square + CF
In the right triangle ABC, ab = AC, angle a = 90 degrees, point D is any point on BC, DF is perpendicular to AB, De is perpendicular to AC, and M is the midpoint What triangle MEF is
It's an isosceles right triangle
Connecting am, according to the meaning of the title, am is the height of BC side in the right triangle ABC, that is ∠ ame + ∠ EMC = 90. And am = BM = MC
It is not difficult to prove that AE = DF = BF, ∠ B = ∠ MAC = 45, so △ BFM ≌ △ AEM, so FM = em, ∠ BMF = ∠ ame,
Therefore, ∠ BMF + ∠ EMC = 90, so ∠ EMF = 90,
So △ MEF is an isosceles right triangle
In the right triangle ABC, ∠ C = 90 degrees, AC = 10cmbc = 5cm, a line segment PQ = ab If two points of PQ move on the line AC and the ray am passing through point a and perpendicular to AC respectively, where does point P move to AC, are △ ABC and △ Apq congruent? If two points of PQ move on the line AC and the ray am passing through point a and perpendicular to AC respectively, what are the P points that meet the conditions to make △ ABC △ def congruent
The first question of this question can be solved, but the second question how to get △ DEF is a little unclear, is it not clear that the problem design is not clear? The solution of the first question is like this, on the AC side of the right triangle, move point P to the distance from point a is 5cm, at this time, △ ABC and △ Apq are congruent
As shown in the figure, in the right triangle ABC, the angle c = 90 degrees, AC = 6cm, BC = 8cm, point PQ As shown in the figure, in the right triangle ABC, the angle c = 90 degrees, AC = 6cm, BC = 8cm, points P and Q start from two points a and C at the same time and move along AC and CB directions at a constant speed. Their speed is 1cm per second. When point P reaches point C, the motion of P and Q stops (1) A few seconds after departure, the area of the triangle PCQ is 4 square centimeters? (2) Can the PCQ area of triangle be 5 square centimeter during the whole movement? Please explain the reason
(1) After setting other T seconds, AP = x, PC = 6-x,
CQ＝x,
S△PCQ＝PC*CQ/2＝4,
（6－t）*t＝8
（t－2）（t－4）＝0,
So when t = 2 or 4 seconds, s = 4
(2) By S = (6-T) * t / 2 = 5
t^2-6t+10=0,
From △ = 6 ^ 2-4 * 10 = - 4 is less than 0,
Or S = (6-T) * t / 2
=（-1/2）（t^2-6t+9)+9/2
=(-1/2)(t-3)^2+9/2
The maximum value of S is 9 / 2. It can't be 5
Given the right triangle ABC, angle c = 90 degrees. AC = BC, P, Q on AB and AP * AP + BQ * BQ = PQ * PQ. Find the angle PCQ
So AP = be, angle CBE = angle a, angle ACP = angle BCE, PC = EC
(1 / 3) in the RT triangle ABC, the angle B = 90 degrees, ab = 5 cm, BC = 7 cm, point P starts from a and moves along the edge of AB at the speed of 1cm / s, and point Q (1 / 3) in the RT triangle ABC, the angle B = 90 degrees, ab = 5 cm, BC = 7 cm, point P moves from a to B at a speed of 1 cm / s, and point Q starts from point B along the edge of BC
Let the pass time be x seconds,
BP = 5-x cm
BQ = 2x cm
(5-X)*2X*1/2=4
X^2-5X+4=0
X1=4 X2=1
After one or four seconds, the area of the triangle PBQ is equal to 4cm ^ 2
(5-X)^2+(2X)^2=25
5X(X-2)=0
X1 = 0, omit x2 = 2
After 2 seconds, the length of PQ is equal to 5cm
As shown in the figure, in the triangle ABC, the angle B = 90 degrees, ab = 6cm, BC = 8cm, point P starts from point a and moves along the edge of AB at a speed of 1cm / s Point Q starts from point B and moves along BC to C at a speed of 2 cm / s (1) P q starts from a and B at the same time. After a few seconds, the s triangle PBQ is 8 square centimeters (2) P and Q start from a and B at the same time, and P to B continue to move on the BC side, and Q to C continue to move on the CA side. After a few seconds, the s triangle PBQ = 12.6 square cm
According to the meaning of the question, set T seconds so that the area of the triangle PCQ is equal to 12.6 square centimeters. Point P moves from point a to side B at the speed of 1cm / s. The total time spent on edge AB is 6S. Point Q moves from point B to point C at the speed of 2cm / s
As shown in the figure, in △ ABC, ∠ B = 90 °, ab = 6cm, BC = 3cm. Starting from point a, point P moves along the edge of AB to point B at the speed of 1cm / s, If P and Q start from a and B at the same time, after a few seconds, the distance between P and Q is equal to 4 times the root sign of 2 cm?
If the moving time is t s, then AP = 6-tcm BQ = 2T cm (0s ≤ t ≤ 3S)
According to the Pythagorean theorem (6-T) ^ 2 + (2t) ^ 2 = 32, t = 2S or T = 0.4s is obtained
RELATED INFORMATIONS
- 1. As shown in the following figure, in Δ ABC, ∠ B = 90 ° and point P moves from point a to point B at the speed of 1cm / s along the edge of AB, and point Q starts from point B and moves along the edge of BC to point C Cm / sec.) (1) If P and Q start from a and B at the same time, and P to B continue to move on the BC side, and Q to C continue to move on the CA side. After a few seconds, the area of Δ PCQ is equal to 12.6 cm? 2 Point Q starts from point B and moves along the edge of BC towards point C at a speed of 2 cm / s
- 2. As shown in the figure, in △ ABC, ab = 6cm, BC = 8cm, ∠ B = 90 °, point P starts from point a and moves along side AB at a speed of 1cm / s, and point Q starts from point B along the edge of BC Point C moves at a speed of 2 cm / s 1) If the points P and Q start from a and B at the same time, after a few seconds, the area of △ PBQ is equal to 8cm? 2) If points P and Q start from a and B at the same time, and P reaches point B and then continues to move on edge BC, and Q reaches point C and then continues to move on edge Ca, after a few seconds, the area of △ PCQ is equal to 9cm?
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- 5. In the right triangle ABC, the angle c = 90 degrees, AC = 4, BC = 3, ab = 5, P is the intersection point of the bisector of the inner angle of the triangle ABC. Find the distance between P and each side
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- 7. As shown in the figure, △ ABC, ab = AC, point E is on the extension line of Ca, and ∠ AEF = ∠ AFE. What is the position relationship between the straight line EF and BC? Why?
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- 10. It is known that: as shown in the figure, ∠ ACB = 90 ° in △ ABC, CD is high, CE bisects ∠ BCD, and ∠ ACD: ∠ BCD = 1:2, then CE is the center line on AB side, right? Give reasons
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- 12. If the ratio of the three inner angles of a triangle is 9:4:5, then the triangle is ()
- 13. In △ ABC, ∠ a = 1 / 3 ∠ B = 1 / 5 ∠ C, calculate the degree of each internal angle of a triangle
- 14. Given that the ratio of the degrees of the three inner angles of a triangle is 4:1:1, then the triangle is () triangle 1. Right angle 2, acute angle 3, obtuse angle
- 15. In the acute angle △ ABC, if the degrees of the three inner angles are prime numbers, then the triangle () A. There is only one isosceles triangle B. There are at least two isosceles triangles C. There is only one, but not an isosceles triangle D. There are at least two of them, including non isosceles triangles
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- 19. In the triangle ABC, the angle a plus the angle B is equal to the angle c, and the angle B is equal to twice the angle A. find the degree of the angle a, the angle B and the angle C
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