As shown in the figure, in △ ABC, ∠ B = 90 °, ab = 3, AC = 5, fold △ ABC so that point C coincides with point a, and the crease is De, then the perimeter of △ Abe is______ ．

As shown in the figure, in △ ABC, ∠ B = 90 °, ab = 3, AC = 5, fold △ ABC so that point C coincides with point a, and the crease is De, then the perimeter of △ Abe is______ ．

∵ in △ ABC, ∵ B = 90 °, ab = 3, AC = 5, ∵ BC = ac2 − AB2 = 52 − 32 = 4, ∵ ade is formed by △ CDE folding, ∵ AE = CE, ∵ AE + be = BC = 4, ∵ Abe perimeter = AB + BC = 3 + 4 = 7

As shown in the figure, in the triangle ABC, D and E are on AC and BC respectively, with angle c = 20, angle CDE = 120 and angle b-4o. Is de and ab even

Parallel

It is known that the images of the first-order functions y = 2x + A and y = - x + B pass through a (- 2,0) and intersect with the Y axis at two points B and C respectively. The area of △ ABC is calculated
Why is B (0,4) and C (0, - 2) processes

Because the images of the linear functions y = 2x + A and y = - x + B pass through a (- 2,0),
So, {- 4 + a = 0
2+b=0
The solution is a = 4, B = - 2
Because the graph of the first-order functions y = 2x + A and y = - x + B intersects with the y-axis at two points B.C
So the coordinates of point B and C are (0,4) (0, - 2) respectively
In conclusion, s △ ABC = [4 - (- 2)] * 2 / 2 = 6

It is known that the images of the first-order functions y = 3-2x + m and y = - 1-2x + Q pass through a (m, 0) and intersect with y axis at B and C 1 respectively to find the area of △ ABC
most urgent!
(2) D is one of the points. The parallelogram is enclosed by points a, B, C and D. write the coordinates of D directly

∵ the images with y = 3 / 2x + m and y = - 1 / 2x + Q pass through a (m, 0)
3 / 2m + M = 0: M = 0
-1 / 2m + q = 0 leads to q = 0
The analytic formula of two straight lines is y = 3 / 2x
y=-1/2x
And ∵ y = 3 / 2x has an intersection with the y-axis, which is in contradiction with your question. Your question may have a problem
And

If the image of a linear function y = - 4 / 3x + 3 intersects the x-axis at point a, and intersects the y-axis at point B. Q: is there a point C on the x-axis such that △ ABC is an isosceles triangle

Let x = 0, y = 3, then B (0,3) let y = 0, x = 9 / 4. Then a (9 / 4,0) if Ba = BC, then the coordinate of point C (- 9 / 4,0) if CB = Ca, let point C (a, 0) 9 / 4-A = √ [A & sup2; + (0-3) & sup2;] 81 / 16-9a / 2 + A & sup2; = A & sup2; + 99A / 2 = - 63 / 16A = - 7 / 8, then point C (- 7 / 8,0) if AC

If the first-order function y = 43x + 4 intersects the x-axis and y-axis at two points a and B respectively, and takes a point on the x-axis so that △ ABC is an isosceles triangle, then such a point C has at most______ One

In y = 43x + 4, let y = 0, the solution is x = - 3; let x = 0, the solution is y = 4. Then the intersection points a and B of the line with X axis and Y axis are (- 3,0), (0,4) respectively. When AB is the bottom edge, vertex C is the intersection point of the vertical bisector of line AB with X axis. When AB is the waist, there are two cases: (1) when a is the vertex of vertex angle

The first-order function y = 4 / 3x + 4 intersects the x-axis respectively. The y-axis is a point on the x-axis at points a, B and C, and the triangle ABC is an isosceles triangle. The coordinates of point C are obtained

A(-3,0)B(0,4),C(X,0)
1)AB=AC
|X+3|=5,C(2,0),(-8,0)
2)AB=BC
X^2+4^2=(-3)^2+4^2,C(3,0)
3)AC=BC
(X+3)^2=X^2+4^2,C(7/6,0)
Coordinates of point C
(2,0)(-8,0)(3,0)(7/6,0)

In △ ABC, ∠ ACB = 90 °, CD ⊥ AB, e is the midpoint of AC, ED.CB When the extension line intersects at one point F, the proof is ac * DF = BC * CF

From CD ⊥ AB, e is the midpoint of AC, de = AE = AC / 2,
So, angle a = angle ade,
Because angle QDE = angle BDF, angle a = angle BDF
From CD ⊥ AB, ∠ ACB = 90 ° angle a = angle BCD,
So, angle BCD = angle BDF
Angle FBD = angle BCD + angle CDB = angle BDF + angle CDB = angle CDF, and angle f is the common angle,
Therefore, triangle CDF is similar to triangle DBF,
So CD: BD = CF: DF
From CD ⊥ AB, ∠ ACB = 90 ° it can be concluded that triangle ABC is similar to triangle CBD
So, AC: CD = BC: BD, that is, AC: BC = CD: BD,
So, AC: BC = CF: DF
That is ac * DF = BC * CF

As shown in the figure, ad is the angular bisector of △ ABC, and the vertical bisectors of ad intersect each other AB.AC At two points of N.M., nd ‖ AC

Proof: as shown in the figure
Mn is the vertical bisector of AD
∴NA=ND
Then, nad = NDA
∵ ad bisection ∠ bac
∴∠NAD=∠CAD
Then, NDA = CAD
∴ND∥AC

It is known that: as shown in the figure, in △ ABC, ab = AC, ∠ a = 120 ° and the vertical bisector Mn of AB intersects BC and ab at points m and N respectively

Proof 1: as shown in the answer figure, connecting am, ∵ - BAC = 120 °, ab = AC, ∵ - B = - C = 30 °, ∵ Mn is the vertical bisector of AB, ∵ BM = am, ∵ - BAM = - B = 30 °, ∵ MAC = 90 °, ∵ cm = 2am, ∵ cm = 2bm