As shown in the figure, it is known that O is a point in the equilateral triangle △ ABC, and the degree ratio of ∠ AOB, ∠ BOC, and ∠ AOC is 6:5:4. In the triangle with OA, ob, and OC as sides, the degree of the angle opposite these three sides is___ ．

As shown in the figure, it is known that O is a point in the equilateral triangle △ ABC, and the degree ratio of ∠ AOB, ∠ BOC, and ∠ AOC is 6:5:4. In the triangle with OA, ob, and OC as sides, the degree of the angle opposite these three sides is___ ．

∫ AOB + BOC + AOC = 360 ° and ∫ AOB: ∫ BOC: ≌ AOC = 6:5:4, ∫ AOB = 144 °, BOC = 120 ° and AOC = 96 °, rotate △ AOC clockwise about point a for 60 ° to get triangle Ao ′ B, connect OO ′, ∫ Δ Ao ′ B ≌ AOC, ≌ Ao ′ B = ≌ AOC = 96 °, O ′ B = OC, Ao

O is a point in △ ABC, and the distance from O to AB, BC and AC of △ ABC is OE = od = of. If ∠ a = 70 °, then ∠ BOC=_____ ?

∠BOC=180-(∠OBC+∠OCB)
=180-1/2(∠B+∠C)
=90+1/2(180-∠B-∠C)
=90+1/2∠A
=125

As shown in the figure, the perimeter of the right triangle AOB is 100, and there are six small right triangles inside it,
Then the sum of the circumference of the six small triangles is______ .
(the easier and more popular the explanation, the better)

The sum of left right sides of six small right triangles is just Ao, the sum of right right right sides of small right triangles is just Bo. The sum of hypotenuse of six small right triangles is just AB, so the sum of perimeter of six small triangles is 100

As shown in the figure, in the RT triangle ABC, the angle BAC = 90, AC = 2Ab, and point D is the midpoint of AC. place a right triangle plate with an acute angle of 45 as shown in the figure

As shown in the figure, in RT △ ABC, the angle BAC = 90 °, AC = 2Ab, and point D is the midpoint of AC. place a right triangle with an acute angle of 45 ° as shown in the figure, so that the two ends of the bevel of the triangle plate coincide with a and D respectively, connecting be and EC
(1) Verification: be = CE
(2) Verification of be ⊥ EC
Is that the problem

In angle ABC, BP bisects angle abc.cp Bisector angle ACB. Angle A56 ° to find the degree of angle BPC
There's no way to put it out. It's about a triangle. The bottom left is B, the bottom right is C, and the top is a. there are bisectors of angle ABC and angle ACB in the middle, and their intersection point is d. if there's a way, please solve it

Angle BPC = 180-1 / 2 (angle ABC + angle ACB) = 180-1 / 2 (180 angle a) = 180-1 / 2 (180-56) = 118

As shown in the figure, in △ ABC, BP and CP are bisectors of ∠ ABC and ∠ ACB, respectively, ∠ BPC = 134 ° and the degree of ∠ A is calculated

In △ BPC, ∠ BPC = 134 °, ≠ 1 + ∠ 2 = 180 ° - ∠ BPC = 180 ° - 134 ° = 46 °, ∵ BP and CP are angular bisectors of ∠ ABC and ∠ ACB, respectively, ≠ ABC = 2 ∠ 1, ∠ ACB = 2 ∠ 2, ≠ ABC + ∠ ACB = 2 ∠ 1 + 2 ∠ 2 = 2 (∠ 1 + ∠ 2) = 2 × 46 ° = 92 °, in △ ABC, ∠ a = 18

In the triangle ABC, the midpoint P is any point BP CP in ABC, which bisects the angle ABC and ACB respectively. Given the angle a = n degrees, the degree of BPC is proved

∵ BP CP bisects the angles ABC and ACB respectively
∴∠PBC=1/2∠ABC,∠PCB=1/2∠ACB
∴∠BPC=180-∠PBC-∠PCB
=180-1/2(∠ABC+∠ACB)
However, ABC + ACB = 180 - A
That is, BPC = 180-1 / 2 (180 - a)
=90+1/2∠A
=90+1/2n

In the triangle ABC, the points D.E are on the sides BC and AC respectively, and | BD | = 1 / 3 | BC |, | CE | = 1 / 3 | Ca |, ad and be intersect at the point P. it is proved that RD = 1 / 7ad

If EF ‖ ad is crossed with BC in F, then
EF/AD=CE/CA=1/3,
∴EF=AD/3,
Similarly, CE = CD / 3,
∴BD=BC/3=(3/7)BF,
∴PD/EF=BD/BF=3/7,
∴PD=(3/7)EF=(1/7)AD.

In triangle ABC, if D is on BC and E is on AC, connect AD and be, BD: BC = 1:3, Ce: CA = 1:3, find DF: FA

Let f be the intersection of AD and be. Let DF = λ dabf-bd = λ (ba-bd) ｜ BF = λ Ba + (1 - λ) BD and BF = μ be = μ (BC + CE) = μ (BC + Ca / 3) ｜ μ BC + μ Ca / 3 = λ Ba + (1 - λ) BC / 3 μ BC + μ Ca / 3 = λ (BC + Ca) + (1 - λ) BC / 3

In RT triangle ABC, angle ACB = RT angle, DC = 4, BD = 2, CD perpendicular to AB, ab

In RT △ BDC
BC²=DC²+BD²=20
BC=2√5
In RT △ BDC and RT △ BCA, ∠ B is the common angle
∴ Rt△BDC∽Rt△BCA
Then AB: BC = BC: BD
That is ab = BC * BC / BD = (2 √ 5) &# 178 / 2 = 10