In the triangle ABC, ab = AC, the line passing through a vertex intersects the opposite side of D. the line divides the original triangle into two isosceles triangles In the triangle ABC, ab = AC, the line passing through a vertex intersects the opposite side of D. the line divides the original triangle into two isosceles triangles. Find the degree of the three inner angles of the triangle ABC.

Think of these two situations first
(1) An isosceles right triangle with a hypotenuse perpendicular from the apex of the right angle
Ninety
Forty-five
Forty-five
(2) Make a line from the top of a base corner
Thirty-six
Seventy-two
Seventy-two

The straight line passing through one vertex of an isosceles triangle divides the two triangles into isosceles triangles. Try to find the degree of each angle of the isosceles triangle There are four situations!

1、 For the acute angle isosceles triangle ABC, take point D on the side of AC so that ad = BD = BC
Let the angle a be X,
Then the angle BDC = 2x
The outer angle of a triangle is equal to the sum of the two corners that are not adjacent to it,
Angle c = angle BDC = 2x,
Angle ABC = angle c = 2x
So x + 2x + 2x = 180
X=36,2X=72
The angles of isosceles triangle are 36 degrees, 72 degrees and 72 degrees respectively
2、 An isosceles right triangle ABC. Angle a is a right angle. Make ad through point a and perpendicular to BC
Triangle abd and triangle ACD are isosceles triangles
Then the angles are 90 °, 45 ° and 45 ° respectively
3、 The obtuse triangle ABC, take point D on BC so that ad = CD, ab = BD
Then angle B = angle c = x, angle BDA = angle bad = 2x, angle DAC = X
So x + X + 2x + x = 180, x = 36
Vertex angle 3x = 108
The angles are 36 ° 36 ° and 108 ° respectively
There are only three situations, acute angle, right angle and obtuse angle. I can't find it

Is there such an isosceles triangle, which is divided into two smaller triangles by a line passing through a vertex, which are also isosceles triangles?

There are. For example: isosceles right triangle, right angle bisector divides it into two identical right isosceles triangle

Divide the isosceles triangle into two isosceles triangles by a straight line passing through one vertex of the isosceles triangle. Try to find out the degrees of each angle of the isosceles triangle (at least three different conclusions are obtained) Divide the isosceles triangle into two isosceles triangles by a straight line passing through one vertex of the isosceles triangle. Try to find out the degrees of each angle of the isosceles triangle (at least three different conclusions are obtained) emergency

1) Vertex a, ab = AC
Make a dividing line to BC through the vertex, so that ab = ad, intersect BC at D
Then ∠ d = (180 - ∠ b) / 2 -- - 1, where ∠ B = ∠ C
∠D=2∠C -------------2
∴2∠C=（180-∠C）/2
∴4∠C=180-∠C
∴5∠C=180
∴∠C=36
∠B=36
∠A=108
2) Similarly, vertex a, ab = AC
Make a dividing line to BC through the vertex, so that ad = BD = DC, crossing BC to d
This is a special case of the RT triangle
∠A=90
∠B=45
∠C=45
3) Vertex a, ab = AC
Make a dividing line to AC through point B, so that ad = CD, intersect BC at D
Then ∠ B = ∠ d = 2 ∠ a -------- 1
∠A=180-2∠B-----------2
∴∠B=2（180-2∠B）
5∠B=360
∠B=72
∠C=72
∠A=36

It is known that: △ ABC is an isosceles triangle. A line passing through its vertex divides △ ABC into two small isosceles triangles, and calculates the degrees of each angle of △ ABC Bonus points for all four kinds of writing

1.180/7 540/7 540/7
2.36 72 72
3.108 36 36
4.90 45 45

The two triangles divided into two isosceles triangles are isosceles triangles. Try to find the degrees of the angles of the isosceles triangle There are four situations

36, 72, 72 -- 36, 36, 108 and 36, 72, 72
108, 36, 36 --- 36, 36, 108 and 36, 72, 72
108, 36, 36 --- 36, 72, 72 and 36, 36, 108
45,45,90 ------ 45,45,90 and 45,45,90

The isosceles triangle ABC with base BC is divided into two smaller isosceles triangles by a straight line passing through a vertex. Please draw all the qualified △ ABC sketches (and mark the degrees of the relevant angles)

For right isosceles triangles, except 90 degrees, it is 45 degrees
For non right isosceles triangles: ∠ 1 = ∠ 2 = ∠ 5, ∠ 3 = ∠ 4
180°=∠1+∠2+∠4+∠5=3∠2+∠4
In ⊿ ADC, 180 ° = ∠ 3 + ∠ 4 + ∠ 5
Therefore, 3 ∠ 2 + ∠ 4 = 2 ∠ 4 + ∠ 2
2∠2=∠4
Finally, ∠ 2 = 36 degrees
∠1=∠2=∠5=36°,∠3=∠4=72°
∠ADE=108°

An isosceles triangle ABC with a base of BC is divided into two small isosceles triangles by a line passing through a vertex There are four kinds of teaching handlebars for urgent use

“wzy85015651”：
The triangle must be an isosceles right triangle, the vertex angle a is a right angle, and the two triangles divided are also isosceles right triangles
Good bye

It is known that the triangle ABC is an isosceles triangle. If a straight line is drawn from its vertex, it will be divided into two new isosceles triangles Waist triangle. To specific steps

There are two such triangles that satisfy the question
In this case, the base angle can be set as X, then the top angle is 2x, so 4x = 180
X = 45 ° is an isosceles right triangle
2 use one base angle of a triangle as the top angle of one of the new triangles, and the other base angle as the base angle of another new triangle
The top angle of the original triangle is 108 and the base angle is 36.72, and the other is 108.36 36

In △ ABC, ab = AC, the line passing through one of the vertices intersects with the opposite side of point D. if it is divided into two triangles, both of them are isosceles triangles Degrees of

Since the shape of the isosceles triangle is not specified in the question, it should be analyzed in four cases to get the answer
Find the top angle, but if the base angle is (180 top angle) / 2, it will not be written
(1) As shown in the figure, △ ABC, ab = AC, BD = ad, AC = CD, calculate the degree of ∠ BAC
∵AB=AC,BD=AD,AC=CD,
∴∠B=∠C=∠BAD,∠CDA=∠CAD,
∵∠CDA=2∠B,
∴∠CAB=3∠B,
∵∠BAC+∠B+∠C=180°,
∴5∠B=180°,
∴∠B=36°,
∴∠BAC=108°．
(2) As shown in the figure, △ ABC, ab = AC, ad = BD = CD
∵AB=AC,AD=BD=CD,
∴∠B=∠C=∠DAC=∠DAB
∴∠BAC=2∠B
∵∠BAC+∠B+∠C=180°,
∴4∠B=180°,
∴∠B=45°,
∴∠BAC=90°．
(3) As shown in the figure, △ ABC, ab = AC, BD = ad = BC, calculate the degree of ∠ BAC
∵AB=AC,BD=AD=BC,
∴∠B=∠C,∠A=∠ABD,∠BDC=∠C
∵∠BDC=2∠A,
∴∠C=2∠A=∠B,
∵∠A+∠ABC+∠C=180°,
∴5∠A=180°,
∴∠A=36°．
(4) As shown in the figure, △ ABC, ab = AC, BD = ad, CD = BC, calculate the degree of ∠ BAC
Suppose ∠ a = x, ad = BD,
∴∠DBA=x,
∵AB=AC,
∴∠C= 180-x/7=2,
∵CD=BC,
∴∠BDC=2x=∠DBC=180-x/7 -x,
The solution is: x = 180 / 7
∴∠A=180/7 ．
Comments: this question mainly investigates the properties of isosceles triangle, the properties of triangle outer angle and the comprehensive application of triangle interior angle sum theorem