Given that circle O passes through the three vertices of triangle ABC, ab = AC, the distance between the center of circle O and BC is 3, and the radius of circle is 7, find the length of ab

1 / 2 * BC = radical (7? - 3?) = heel 40
AB = root number ((root number 40) 2 + (7 + 3) 2) = 2 root number 35
Or AB = root number ((root number 40) 2 + (7-3) 2) = 2 root number 14

It is known that the circle O passes through the three vertices of △ ABC, and ab = AC, the distance between the center O and BC is 3, the radius of the circle is 7, and the length of the ball ab

Connected to ob, through o as OD ⊥ BC
In the right triangle OBD, from the Pythagorean theorem, it is concluded that,
BD^2=BO^2-OD^2=7²-3²=40
In the right triangle abd,
From the Pythagorean theorem, we get that,
AB²=AD²+BD²=(7+3)²+40=140
AB = 2 √ 35

In the triangle ABC, ab = AC, circle O passes through three points of ABC with radius of 4, and the distance from point O to BC is 1. Find the length of ab

Connect OA to BC to E
When above, AE = oa-oe = 4-1 = 3
BE=√OB²-OE²=√15
AB=√AE²+BE²=√21
When it is below, AE = OA + OE = 5
BE=√OB²-OE²=√15
AB=√AE²+BE²=2√10

In the isosceles triangle ABC, ab = AC = 2cm. If the circle with a as the center and 1cm as the radius is tangent to BC, calculate the degree of angle ABC Please explain the reason

Through a, make ad ⊥ BC
Ad is the radius of 0 a
In RT △ ADB
∠ADB=90°.
∵AB=2cm,AD=1cm
∴cos∠DAB=AB/AD
ν angle DAB = 60 °
Three lines of isosceles triangle are combined into one, and the angle ABC = 120 ° is obtained

As shown in the figure, △ ABC is an isosceles triangle, ab = AC, O is the midpoint of the bottom BC, ⊙ O and waist AB are tangent to point D. It is proved that AC is tangent to ⊙ o

It is proved that: connect OD, and make OE ⊥ AC at point E,
Then ∠ OEC = 90 °,
∵ AB cut ⊙ o in D,
∴OD⊥AB，
∴∠ODB=90°，
﹤ ODB = ∠ OEC; (3 points)
Again, O is the midpoint BC,
∴OB=OC，
∵AB=AC，
∴∠B=∠C，
≌△ OCE, (6 points)
⊙ OE = OD, that is, OE is the radius of ⊙ o,
⊙ AC is tangent to ⊙ O. (9 points)

As shown in the figure, △ ABC is an isosceles triangle, ab = AC, O is the midpoint of the bottom BC, ⊙ O and waist AB are tangent to point D. It is proved that AC is tangent to ⊙ o

Given that the waist ab of the isosceles triangle ABC is 4, if the circle with a as the center and 2 as the radius is tangent to BC, then BC is equal to

solution
Because it is tangent to BC, the distance from a to BC is 2
So 1 / 2BC = √ AB ^ 2-2 ^ 2 = 2 √ 3
So BC = 2 √ 3 * 2 = 4 √ 3

If the isosceles △ ABC is inscribed on ⊙ o with radius 5, and the distance between point O and bottom BC is 3, then the length of AB is___ .

Consider two cases: when △ ABC is an acute triangle, as shown in Fig. 1, through a as ad ⊥ BC, from the meaning of the title, ad is over the center O, connecting ob, ∵ od = 3, OB = 5,  in RT △ OBD, according to the Pythagorean theorem, BD = 4; in RT △ abd, ad = Ao + od = 8, BD = 4; according to the Pythagorean theorem, ab = 82 + 42 =

If we join the triangle ABC in the circle so that ab = AC, the radius of the circle is 7cm, and the distance from the center of the circle to BC is 3cm, then AB =?

If the inscribed triangle ABC, ab = AC, then point a, center O, midpoint of BC (set as d) are on the same line, and ad is perpendicular to BC
Proof process is very simple, do not write
BD = 10 root triangle
Abd is a right triangle, ab = 140 under radical sign = 35 under 2 radical sign

The triangle ABC is inscribed in the circle O, ab = AC, BC = 6, the distance from point O to BC is 4, find ab

BC = 6 o and the distance from O to BC is 4
Because AB = AC, according to the isosceles triangle three lines in one, we know that ad is also the center line
That is, ad is the vertical bisector of string BC, then ad must pass through the center o
Connect ob
In right angle △ BOD
∵OD=4,DB=3
∴OB=5
In the right angle △ abd,
∵AD=5+4=9,BD=3
∴AB=3√10

Given that circle O passes through the three vertices of triangle ABC, ab = AC, the distance between the center of circle O and BC is 3, and the radius of circle is 7, find the length of ab