As shown in the figure, in △ ABC, BD and CD divide ∠ ABC and ∠ ACB equally, and try to explain ∠ d = 90 ° + & # 189; ∠ a

As shown in the figure, in △ ABC, BD and CD divide ∠ ABC and ∠ ACB equally, and try to explain ∠ d = 90 ° + & # 189; ∠ a

prove:
∵∠A+∠ABC+∠ACB＝180
∴∠ABC+∠ACB＝180-∠A
∵ BD bisection ∠ ABC
∴∠DBC＝∠ABC/2
∵ CD bisection ∠ ACB
∴∠DCB＝∠ACB/2
∴∠D＝180-（∠DBC+∠DCB）
＝180-（∠ABC/2+∠ACB/2）
＝180-（∠ABC+∠ACB）/2
＝180-（180-∠A）/2
＝90+∠A/2

It is known that CD is the height of triangle ABC, angle ACB = 90 degrees, please cooperate with your partner to explain the reason why angle a = angle BCD

The angle B is shared by two triangles ABC, and the angle BDC = ACB = 90 °, so the angle a = B

Given that the angle AOB is equal to 90 ° and OC is any ray outside ∠ AOB, OM bisects ∠ AOC and on bisects ∠ BOC, and the degree of ∠ mon is obtained
It's better to answer with ∵

Solution
∵ on bisection ∠ BOC
∴∠bon=1/2*∠boc
∴∠aon=∠aob+∠bon
∵∠aob=90°
∴∠aoc=90+∠boc
∴∠aon=90+∠bon
∵ om bisection ∠ AOC
∴∠aom=1/2*∠aoc=45°+1/2*∠boc=45°+∠bon
∴∠mon=∠aon--∠aom=(90°+∠bon)--(45°+∠bon)=45°

The angle AOB and the ray OC, OM and on are known to divide the angle AOC and the angle BOC equally. (1) if OC is outside the angle AOB, try to explore

The angle AOB and the ray OC, OM and on are known to divide the angle AOC and the angle BOC equally. If OC is outside the angle AOB, the relationship between the angle mon and the angle AOB is studied

It is known that: ∠ mon and a point a in ∠ mon can be calculated as △ ABC, so that B is on OM, C is on, and the perimeter of △ ABC is the smallest. (the figure is relatively simple. I'll trouble you.)

Take OM and on as the axis of symmetry, make two symmetrical points of a, let them be a 'and a "OK, connect a' and a", there will be two intersections with OM and on, let them be B and C, and then connect AB, BC and AC

A is a point in ∠ mon. Try to make a point B and C on the OM and on sides to minimize the perimeter of triangle ABC, and explain the reason
How to draw a picture? Process, reason, thank you

Method:
1. Make a point of symmetry A1 and A2 about Mo and no respectively
2. Connect A1, A2, OM to B, on to C
3. Connect ABC, △ ABC is the result
I can't give you a picture. Draw by yourself
The simple proof is as follows:
BA1 = Ba, Ca2 = Ca, ∵ A1, A2 in a straight line, ∵ C △ ABC is the shortest

In the triangle ABC, a (5,2) B (7,3) and the midpoint m of AC is on the y-axis, and the midpoint n of BC is on the x-axis, the coordinates of vertex C and the equation of line Mn are obtained
There are two questions in this question. I hope the more detailed the answer, the better,

Tell me, did you get a (5, - 2) wrong
The method is as follows
1. Let the coordinates (x, y) of C be 5 + x = 0 because the midpoint m of AC is on the Y axis,
X = - 5, and the midpoint n of BC is on the X axis, so y + 3 = 0, y = - 3
So C (- 5, - 3)
2. Because C (- 5, - 3), we get m (0, - 5 / 2), n (1,0) from the title
Let the equation of the line Mn be y = KX + B,
So - 5 / 2 = B, 0 = K + B, so k = 5 / 2, B = - 5 / 2
So the equation of the line Mn is y = 5 / 2x-5 / 2

In the triangle ABC, a (- 1,2), B (4,3), and the midpoint m of edge AC is on the y-axis, and the midpoint of edge BC is on the x-axis

Let C (a, b), then the midpoint of AC m [(A-1) / 2, B + 2) / 2] in Y-axis (A-1) / 2 = 0, a = 1n [(4 + a) / 2, (3 + b) / 2] in x-axis (3 + B) / 2 = 0b = - 3, so C (1, - 3) so m (0, - 1 / 2), n (5 / 2,0) so Mn slope = (0 + 1 / 2) / (5 / 2-0) = 1 / 5, so y = 1 / 5 (X-5 / 2) x-5y-2 = 0

Given the two vertices a (4.7) B (- 2.6) of triangle ABC, we can find the coordinates of point C such that the midpoint of AC is on the x-axis and the midpoint of BC is on the y-axis

Let the coordinates of point C be (m, n)
Then the midpoint coordinates of AC are (M / 2 + 2, N / 2 + 7 / 2)
The midpoint coordinates of BC are (M / 2-1, N / 2 + 3)
Because the midpoint of AC is on the x-axis and the midpoint of BC is on the y-axis
So n / 2 + 7 / 2 = 0, M / 2-1 = 0
The solution is n = - 7, M = 2
So the coordinates of point C are (2, - 7)

Given the two vertices a (3,7) and B (- 1,3) of the triangle ABC, find the coordinates of point C so that the midpoint of AC is on the X axis and the midpoint of BC is on the Y axis!

(1,-7)