It is known that △ ABC is an isosceles triangle, BD is high, and ∠ abd = 50 degrees It's a process

There are three answers,
When AB = AC, vertex angle a = 40 ° or 50 °
When CA = CB, vertex angle c = 50 ° or 100 °
When Ba = BC, vertex angle B = 100 °

As shown in the figure, ad is the height of △ ABC, AE is the angular bisector of △ ABC, AF is the middle line of △ ABC, and the equal angles in the figure are The equivalent line segments are ．

∵ AE is the angular bisector of △ ABC, ∵ BAE = ∠ CAE. ∵ ad is the height of △ ABC, ∵ ADB = ∠ ADC = 90 °. ∵ AF is the middle line of △ ABC, ∵ BF = CF

In △ ABC, ab = 15, AC = 20, height ad = 12, AE is an angular bisector, then the length of AE is______ ．

(1) High ad is on the outside of triangle ABC: in right triangle abd, according to Pythagorean theorem: BD = 9, CD = 16, BC = 9 + 16 = 25, ∵ BC2 = 625, AB2 = 225, ac2 = 400, ∵ ac2 + AB2 = BC2 ∵ a = 90, ∵ AE is angular bisector, ∵ BAE = 45 degree, ∵ SINB = 45

As shown in the figure, △ ABC, ab = 10, ∠ B = 2 ∠ C, ad is the high line, AE is the middle line, then the length of line De is______ ．

Through e, make me parallel to ad, cross AC to m, ∵ ad is high line, ∵ ad ⊥ CB, ∵ me ⊥ CB. Connect BM, in △ CBM, me is middle line and high line, ∵ MBE is isosceles triangle, ∵ BM = cm, ∠ C = ∠ CBM, and ∵ B = 2 ∠ C, ∵ MBA = ∠ C, and ∵ cab = ∠ cab, ∵ mAb ∽ BAC, ∵ ABMA = cbmb = CBMC. ∵ me ∥ ad, ∥ ceed = CMMA, CE = 12cb, ∥ CBCM = 2edam, ∥ AB = 2DE So the answer is: 5

The locus of the vertex C of an isosceles triangle with ab as its base is______ ．

∵△ ABC takes line AB as the bottom edge, CA = CB, and the point C is on the vertical bisector of line AB, excluding the intersection point with ab (the intersection point does not meet the triangle condition). The locus of vertex C of isosceles triangle with line AB as the bottom edge is the vertical bisector of line AB, excluding the middle point of ab. so the answer is the vertical bisector of line AB, excluding the middle point of ab

The isosceles triangle with ab = a has_ The vertex that meets the condition is at the end of line ab__

Problem 1: there is no waist length, so there are countless
Question 2: there is no bottom angle, so there are countless

Given the vertex a (3,20) of an isosceles triangle and the vertex B (3,5) of a base angle, we can find the trajectory equation of the other vertex C (x, y)
This mathematics is urgent noise, annoys dead Le, who helps to look down

（x-3）×(x-3)+(y-20)(y-20)=15×15

The vertex a (3,20) and the vertex B (3,5) of an isosceles triangle are known
Find the trajectory equation of another vertex C

Let the coordinates of point C be (x, y)
AC=AB
The trajectory equation of another vertex C
(x-3)²+(y-20)^2=(3-3)²+(20-5)²
(x-3)²+(y-20)²=15²
（x-3）²+(y-20)²=225 (x≠3)

The coordinates of the vertex a of an isosceles triangle are (4,2), and the coordinates of the bottom endpoint B are (3,5). Find the trajectory equation of the other endpoint C

Set point C (x, y)
According to an isosceles triangle, the two waists are equal
AB=AC
Distance formula:
(4-3)^2+(2-5)^2=(x-4)^2+(y-2)^2
Trajectory equation of endpoint C
(x-4)^2+(y-2)^2=10

Given that the coordinates of the two ends of the base of an isosceles triangle are B (4,2) C (- 2,0), the trajectory equation of the third vertex A is obtained
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AB=AC
So a is on the vertical bisector of BC
BC midpoint (1,1)
BC slope (2-0) / (4 + 2) = 1 / 3
So BC vertical bisector slope = - 3
y-1=-3(x-1)
3x+y-4=0
ABC is not collinear
So 3x + y-4 = 0, excluding (1,1)

It is known that △ ABC is an isosceles triangle, BD is high, and ∠ abd = 50 degrees It's a process