As shown in the figure, it is known that in △ ABC, ab = AC, D is the point on AB, de ⊥ BC, e is the perpendicular foot, the extension line of ED intersects the extension line of Ca at point F, and the proof is: ad = AF

As shown in the figure, it is known that in △ ABC, ab = AC, D is the point on AB, de ⊥ BC, e is the perpendicular foot, the extension line of ED intersects the extension line of Ca at point F, and the proof is: ad = AF

It is proved that: ∵ AB = AC, ∵ B = ∵ C, ∵ de ⊥ BC, ∵ C + F = 90 °, B + BDE = 90 °, ∵ ADF = BDE, ∵ f = ADF, ∵ ad = AF

As shown in the figure, in △ ABC, ∠ B = C, points D and E are points on the side of BC, and ∠ C = ∠ DAE, please explain the reason why ∠ BAE = ∠ ade
The sum of the internal angles of a triangle can be 180 or one of the external angles of a triangle is equal to the sum of its two non adjacent internal angles

∠BAE=∠BAD+∠DAE=∠BAD+C
∠ADE=∠B+∠BAD=∠C+∠BAD
That is, BAE = ade

Please solve it as soon as possible. As shown in the figure, in triangle ABC, angle B = angle c, points D and E are points on the side of BC, and angle c = angle DAE. Please explain the reason why angle BAE = angle ade

Angle c = angle DAE, angle ade common angle, so triangle DAE is similar to triangle DCA, so angle DAC = angle DEA, so triangle bad is similar to triangle BEA, so angle BAE = angle ade

Given that the length of one side BC of the triangle ABC is 4, the area is 6, and the vertex a changes, the trajectory equation of the center of gravity g of the triangle ABC is obtained

Take BC as the x-axis to establish the coordinate system, take the midpoint of BC as the center of circle to establish the coordinate system, and let the center of gravity g coordinate be (x, y). Because the area of triangle BCG is one-third × 6 = 2, so one-half times BC times the absolute value of y = 2, so y = 1 (plus or minus 1) - the trajectory equation

If line AB = 3cm, the locus of vertex C of right triangle with line AB as hypotenuse

A circle with the diameter of point AB, but point C cannot coincide with points a and B!

The locus of the right angle vertex C of a right triangle with ab as the hypotenuse is

Make a height from point C to AB, take this height as the radius, and the circle with the intersection of this line and ab as the center is the trajectory of C

Given the vertex a (5,1) of △ ABC, the linear equation of the central line cm on the edge AB is 2x-y-5 = 0, and the linear equation of the high BH on the edge AC is x-2y-5 = 0. (I) find the linear equation of the edge AC; (II) find the coordinates of the vertex C; and;

(I) from the linear equation of high BH on AC side is x-2y-5 = 0, we can know that KAC = - 2, and a (5,1), the linear equation of AC side is Y-1 = - 2 (X-5), that is, the linear equation of AC side is 2x + Y-11 = 0. (II) from the linear equation of AC side is 2x + Y-11 = 0, the linear equation of central line cm on AB side is 2x -

Given the vertex a (5,1) of △ ABC, the linear equation of the central line cm on the edge AB is 2x-y-5 = 0, and the linear equation of the high BH on the edge AC is x-2y-5 = 0

(1) Let C (m, n), ∵ AB be 2x-y-5 = 0 and AC be x-2y-5 = 0. Let 2m-n-5 = 0n-1m-5 × 12 = - 1, then M = 4N = 3. Let C (4, 3); (2) let B (a, b), then a-2b-5 = 02 × a + 52-1 + b2-5 = 0, and a = 1b = 7. Let B (1, 7). Let KBC = 7-31-4 = - 43. Let Y-3 = - 43 (x-4) be changed to 4x + 3y-2 5=0．

Given the linear equation x-2y-5 = 0 where the height BH is on the edge of a (5,1) AC. the linear equation 2x-y = 0 where Zhongxian cm is on the edge of AB, the coordinates of vertex C can be obtained

∵AC⊥BH ∴kac=-1/kbh=-2
The linear equation of AC is y-ya = KAC (x-xa) = > Y-1 = - 2 (X-5) = > 2x + y = 11
And the equation of CM is 2x-y = 0 = > 4x = 11 = > x = 11 / 4
=> y=22/4
The coordinates of point C (11 / 4, 22 / 4)

It is known that the linear equation of vertex a (3,2) of △ ABC, bisector CD of angle c is Y-1 = 0, and the linear equation of high BH on edge AC is 4x + 2y-9 = 0
(1) The coordinates of point C;
(2) The area of the triangle ABC

k(BH)=-2k(AC)=1/2AC:y-2=(1/2)*(x-3)(1)y=1,x=1C(1,1)(2)k(BC)=-k(AC)=-1/2BC:y-1=(-1/2)*(x-1).(1)BH:4x+2y-9=0.(2)B(2,0.5)AC:x-2y+1=0|BH|=|2-2*0.5+1|/√5=2/√5|AC|=√5s=(1/2)*√5*2/√5=1