As shown in the figure, in ABC, three bisectors ad, be, CF intersect at O, og ⊥ BC at g, and prove: BOD = GOC!

∠BOD=∠BAO+∠ABO
=1/2(∠A+∠B)
∠COG=90°-∠OCG
=1/2(180°-∠C)
=1/2(∠A+∠B)
∴∠BOD=∠COG

In the triangle ABC, ad, be and CF are the bisectors of the three inner angles, and they intersect at point O, go through point O, do og, and be perpendicular to g. prove that the angle BOD = angle cog

According to the meaning of the title:
∠BOD=∠BAO+∠ABO
=1/2(∠A+∠B)
∠COG=90°-∠OCG
=1/2(180°-∠C)
=1/2(∠A+∠B)
∴∠BOD=∠COG

In 58 face triangle ABC, three bisectors ad BF CE intersect at point O, og is perpendicular to BC, and ∠ BOD = ∠ GOC is proved
I don't have a graph, but it's just a common triangle and three bisectors of internal angles intersect at O, ABC, where the three points are as follows
A
B C

Let ∠ ABO = α, then ∠ CBO = α, ∠ Bao = θ, then ∠ Cao = θ, ∠ ACO = β, then ∠ BCO = β, ∠ BOD = x, ∠ dog = Z, ∠ cog = y. according to the external angle theorem, the two acute angles of a right triangle are complementary to each other, and then we can get: ① x = α + θ, ② Z + y = β + θ, ③ α + X + Z = 90 ° and ④ y + β = 90 ° respectively

As shown in the figure, the bisectors ad, be and CF of each angle of △ ABC intersect at O, make Ag ⊥ BC at g, and verify: ⊥ BOD = ⊥ cog

∵∠BOD=∠OAB+∠OBA
=（∠ABC+∠BAC）/2
=（180°-∠ACB）/2
=90°-∠ACB/2
=90°-∠OCB
Ψ△ OGC is a right triangle
∵∠GOC=90°-∠OCB
∴∠BOD=∠COG

In △ ABC, ad, be and CF are bisectors of three internal angles respectively, then ∠ 1 ＋ 2 ＋ 3 = ()

What about the picture
But it should be the bisector of each corner
∠1＋∠2＋∠3=1/2(∠A+∠B+∠C)=1/2*180=90

The triangle ABC is an isosceles triangle. The bisector AE of the top angle and the bisector be of the bottom angle intersect at point E
If AE = EF, find the degree of ∠ C

It is proved that a point D on Ba extension line is easy to get ∠ DAE = ∠ EAC = ∠ C
So AE ‖ BC
Therefore, e = EBC
Because be is the bisector of ABC, and C = ABC
Therefore, C = 2 E
I don't seem to know what the F is

As shown in the figure, in △ ABC, the bisector AE of the outer angle of ∠ BAC is parallel to BC, indicating that △ ABC is an isosceles triangle

The bisector AE of BAC is parallel to BC
So these two small external angles are equal to ∠ B and ∠ C respectively (one is equal to the internal stagger angle and the other is equal to the apposition angle)
Therefore, B = C
Then △ ABC is an isosceles triangle

How to prove that the sum of internal angles of triangle ABC is 180 degrees
Prove that the sum of internal angles of triangle ABC is 180 degrees

1. Fold the three corners of a triangle inward. The three corners just form a flat angle, so it is 180 degrees
2. Make the parallel line of the opposite side at a vertex, and prove it with internal stagger angle
three
Do triangle ABC
Make a straight line EF parallel to BC through point a
Angle EAB = angle B
Angle fac = angle c
Angle EAB + angle fac + angle BAC = 180
Angle BAC + angle B + angle c = 180
4. Inner angle sum formula (n-2) * 180
5. Let three vertices of triangle be a, B and C, corresponding to angle a, angle B and angle C respectively; make a straight line L parallel to line BC through point a, make an angle B 'between L and ray AB, make an angle c' between L and ray AC, and make an internal stagger angle between angle B 'and angle B, angle c' and angle C respectively. According to the principle that the internal stagger angles of parallel lines are equal, we can get: sum of internal angles of triangle = angle a + angle B + angle c = angle a + angle B '+ angle c' = 180 degrees
6. Extend the sides of triangle ABC, DAB = C + B, EBA = a + C, FCA = a + B
So DAB + EBA + FCA = 2A + 2B + 2C = 360
So a + B + C = 180
7. Extend one side of a triangle to form the diplomacy of a triangle. It is easy to find that this angle and the internal angle of the triangle adjacent to it add up to a flat angle (180 degrees), so they are adjacent complementary angles. Then make a straight line parallel to the opposite side of the angle through the vertex of the internal angle, and divide the diplomacy into two angles, It can be proved that the other two angles of a triangle are equal to the two angles separated from the triangle. Then the sum of the three inner angles of a triangle is equal to the inner angle plus its adjacent complementary angle, which is 180 degrees
8. Mark three triangles of the same size with three letters a, B and C respectively on the positions of three corresponding angles. Then put together the angle a of the first triangle, the angle B of the second triangle and the angle c of the third triangle, At this time, their lower (or upper) side just forms a straight line. That is to say, the three angles form a flat angle. That is to say, the sum of the degrees of the three angles is 180 degrees. These three angles are the three inner angles of the triangle

It is known that the perimeter of triangle ABC is 16 cm, ad is the middle line on the side of BC, ad = 4 / 5 AB.AD=4cm The circumference of the triangle abd is 12cm
It is known that the perimeter of triangle ABC is 16 cm, ad is the middle line on the side of BC, ad = 4 / 5 AB.AD=4cm The circumference of the triangle abd is 12cm. Draw a diagram in line with the meaning of the question and find the length of each side of the triangle ABC

The triangle ABC, ab = 5cm, AC = 5cm, BC = 6cm solution ad = 5 / 4AB, ad = 4cm can get AB = 5cm, and the circumference of the triangle abd is 12cm, so BD = 3cm, because ad is the midline on the BC side, so DC = 3, the circumference of the triangle ABC is 16cm, so AC = 5

In △ ABC, AC = BC, the center line ad on the side of BC divides the perimeter of triangle into two parts, 12cm and 15cm, and calculates the length of each side of △ ABC
In ten minutes

AC = BC = 10, ab = 7, draw a triangle,
AC = BC = 2x, ab = x-3. The sum of the three is the circumference, so we can get X

As shown in the figure, in ABC, three bisectors ad, be, CF intersect at O, og ⊥ BC at g, and prove: BOD = GOC!