As shown in the figure, in RT △ ABC, ∠ C = 90 °, B = 30 °, BC = 4cm, with point C as the center of the circle and 2cm long as the radius as the circle, then the position relationship between ⊙ C and ab is () A. Separation B. Tangency C. Intersection D. Tangency or intersection

As shown in the figure, in RT △ ABC, ∠ C = 90 °, B = 30 °, BC = 4cm, with point C as the center of the circle and 2cm long as the radius as the circle, then the position relationship between ⊙ C and ab is () A. Separation B. Tangency C. Intersection D. Tangency or intersection

Make CD ⊥ AB at point D
∵∠B=30°，BC=4cm，
∴CD=1
2BC=2cm，
That is, CD is equal to the radius of the circle
∵CD⊥AB，
⊙ AB is tangent to ⊙ C
Therefore, B

In the RT triangle ABC, ∠ C is equal to 90 °, AC is equal to 3cm, BC is equal to 4cm. With C as the center of the circle, what is the position relationship between the circle with radius R and ab

Taking angle c as height, the height length is 4 * 3 / 5 = 2.4 (CM)
When R < 2.4 (CM), AB is outside the circle
When r = 2.4 (CM), AB is tangent to the circle
When 3 > = R > 2.4 (CM), AB intersects the circle at two points
4> When = R > 3 (CM), AB intersects the circle
When r > 4 (CM), AB is in the circle
There are still questions, please ask again!

As shown in the figure, we know the value of AB = AC, ∠ BAC = 120 ° and Qiao AB: BC in the triangle ABC I saw your problem-solving process, but the following BC = root sign 3AB becomes AB: BC = 3 / 3 root sign 3, I don't know how to change it

Let AB = 1, then BC = 2 ×√ 3 / 2 = √ 3
AB：BC=1：√3=√3：3.

It is known that: as shown in the figure, in △ ABC, ab = AC, ∠ a = 120 ° and the vertical bisector Mn of AB intersects BC and ab at points m and N respectively. It is proved that CM = 2bm

Proof 1: as shown in the answer figure, connect am, ∵ BAC = 120 °, ab = AC,  B = ∠ C = 30 °,

It is known that: as shown in the figure, in △ ABC, ab = AC, ∠ a = 120 ° and the vertical bisector Mn of AB intersects BC and ab at points m and N respectively. It is proved that CM = 2bm

Proof 1: as shown in the answer figure, connect am, ∵ BAC = 120 °, ab = AC,  B = ∠ C = 30 °,

0

Proof 1: connect am as shown in the answer,
∵∠BAC=120°，AB=AC，
∴∠B=∠C=30°，
∵ Mn is the vertical bisector of ab,
∴BM=AM，∴∠BAM=∠B=30°，
∴∠MAC=90°，
∴CM=2AM，
∴CM=2BM．
Proof two: as shown in the answer chart, pass a
Make ad ∥ Mn to BC at point D
∵ Mn is the vertical bisector of ab,
ν n is the midpoint of ab
∵AD∥MN，
ν m is the midpoint of BD, i.e. BM = MD
∵AC=AB，∠BAC=120°，
∴∠B=∠C=30°，
∵∠BAD=∠BNM=90°，
∴AD=1
2BD=BM=MD，
And ? CAD = ∠ BAC - ∠ bad = 120 ° - 90 ° = 30 °,
∴∠CAD=∠C，
∴AD=DC，BM=MD=DC，
∴CM=2BM．

As shown in the figure, in the triangle ABC, ab = AC, the angle BAC = 120 ° and the vertical bisector Mn of AB intersects BC, AB at m, n respectively. It is proved that CM = 2bm

∵AB=AC,∠BAC=120°,
According to the theorem of sum of internal angles: ∠ B = ∠ C = 30 °,
Connecting Ma, ∵ Mn is the vertical bisector of the line segment ab,
∴MA=MB,∴∠BAM=∠B=30°,
∴∠MAC=90°,
In the right angle △ AMC,
∵∠C=30°,∴MC=2AM,
∴CM=2BM.

In the isosceles triangle ABC, ab = AC, angle BAC = 120 degrees, the vertical bisector of AB intersects BC at D, and BD = 6 cm, what is BC = then?

Do AB bisection point F, AE is perpendicular to BC and E, because angle BAC=120 degrees, AB=AC, triangle ABC is isosceles triangle, so angle B is (180-120) divided by 2, equal to 30 degrees. In triangle BDF, angle BFD is equal to 90 degrees, angle B is equal to 30 degrees, so DF is equal to 1/2BD, equal to 3, BF square is equal to ad square minus DF square, equal to triple root

In angle ABC, ab = AC, angle ABC. The bisector of angle ACB intersects with O, indicating that angle BOC is an isosceles triangle

Because AB is equal to AC, angle ABC is equal to angle ACB
Because the bisector of angle ABC. Angle ACB intersects with O, angle OBC = angle OCB
Because the angle OBC = the angle OCB, so ob = OC, the angle BOC is an isosceles triangle
(big brother, what you mean by "angle ABC" should be "triangle ABC". It's easy to confuse people's thinking

In the isosceles triangle ABC, the heights BD and CE of the two waists intersect at O. question: is the triangle BOC isosceles triangle? Why?

Yes