The radius of the circle is known to be 4, and a, B and C are the three sides of the inscribed triangle of the circle. If ABC = 16 √ 2, the area of the triangle is () A 2√2 B 8√2 C √2 D √2/2

The radius of the circle is known to be 4, and a, B and C are the three sides of the inscribed triangle of the circle. If ABC = 16 √ 2, the area of the triangle is () A 2√2 B 8√2 C √2 D √2/2

abc=16√2
ab = 16√2/c
S = 1/2absinC = 1/2*16√2/c sinC = 1/2*16√2/(c/sinC)
According to the sine theorem:
a/sinA=b/sinB=c/sinC = 2R = 8
∴S = 1/2*16√2/(c/sinC) = 8√2/8 = √2
C is correct

If the area of the inscribed triangle in a circle of radius 1 is 1, then the product of the three sides of the triangle ABC

First of all, we should know that the inscribed triangle of a circle is a right triangle, and the hypotenuse is the diameter
2, so the area is 1, and the information provided to us is that half of the product of two right angles is 1, then the product of these two right angles is 2
The radius is 1 and the diameter is 2
4, the product of two right angles and bevel sides is 2 * 2 = 4
5. I hope you find more hidden conditions when you encounter problems

4. If the area of inscribed triangle of radius 1 is 1 / 4, then the product of three sides' length ABC=___ ? Please write down the detailed process and ideas

I'm a graduate of this year. I can't think of anything else but use a special case. One side of this triangle is over the center of a circle, so it is a right triangle, and one side is a = 2, BC = 1 / 2 product ABC = 1

As shown in the figure, △ ABC is an acute triangle, with BC as the diameter to make ⊙ o, ad is the tangent line of ⊙ O. from a point E on AB, make the vertical line of AB intersect the extension line of AC at F, if ab AF＝AE AC． Confirmation: ad = AE

It is proved that: as shown in the figure, let AC intersect ⊙ o at point n. connect BN, ∵ BC be the diameter of ⊙ o,

As shown in the figure, △ ABC is inscribed on the point o with angle B = 60 ° and the tangent l of the circle is made through point C and the extension line of diameter ad is educated to point E; AE is vertical to L, CG is vertical to AD If the perpendicular foot is g (1), the triangle is equal to the triangle ACG (2) if

(1) As shown in the figure, if CD and OC are connected, then ∵ AC ⊥ CD, CG ⊥ ad, ? ACG = ∠ ADC = 60 °. Since ∠ ODC = 60 ° OC = OD,  OCD is an equilateral triangle, we can get ﹣ DCO = 60 °. From OC ⊥ L, ? ECD = 30 °, CG ⊥ ad, ? ACG = 30 ° + 30 ° = 60 °

As shown in the figure, △ ABC is connected to ⊙ o, point D is on the extension line of OC, SINB = 1 2，∠CAD=30°． (1) It is proved that ad is tangent of ⊙ o; (2) If OD ⊥ AB, BC = 5, find the length of AD

It is proved that: (1) ∵ SINB = 12,  B = 30 °, AOC = 60 ° and

As shown in the figure, in the triangle ABC, the angle c = 90 degrees, taking BC as the diameter, making the circle intersection ad in D, and making the tangent line of the circle through D intersect AC in E. it is proved that e is the middle of AC As shown in the figure, in the triangle ABC, the angle c = 90 degrees, with BC as the diameter, make the circle intersection ad at D, and make the tangent line of the circle AC intersect e through D. It is proved that e is medium of AC spot

Let the midpoint of AB be o, connect OD, OE and CD, and let CD and OE intersect with F
The angle CDB is the circumference of the circle, and the angle is equal to 90
Right triangle OC = OD, OE = OE
So OCE is equivalent to ODE
So EC = ed, angle OEC = OED
In isosceles triangle ECD, OE is the bisector of the angle CED
So the OE vertical CD,
So EF parallels dB
Triangle ABC similar EOC
CE/AC=OC/BC=1/2
So e is the midpoint of AC

As shown in the figure, it is known that ⊙ o is the circumscribed circle of ⊙ ABC, and ab is the diameter. If PA ⊥ AB, Po passes through the midpoint m of AC, it is proved that PC is the tangent of ⊙ o

Proof: connect OC,
∵PA⊥AB，
﹤ PA0 = 90 °. (1 point)
∵ Po passes through the midpoint of AC, OA = OC,
Ψ Po bisection ∠ AOC
﹤ AOP = ∠ cop. (3 points)
In △ Pao and △ PCO, OA = OC, ∠ AOP = ∠ cop, Po = Po
≌△ PCO. (6 points)
∴∠PCO=∠PA0=90°．
That is, PC is the tangent line of ⊙ O. (7 points)

As shown in the figure, the circle O is the circumscribed circle of △ ABC, and the extension of the tangent intersection ab of point C is at point D, CD = 2 7, ab = BC = 3. Find the length of BD and AC

According to the cut line theorem, DB? Da = DC2, that is dB (DB + BA) = DC2,
DB2 + 3db-28 = 0, DB = 4
∵∠A=∠BCD，∴△DBC∽△DCA，
∴BC
CA=DB
DC，
AC = BC · DC
DB=3
Seven
2．

It is known that: as shown in the figure, in △ ABC, D is a point on the edge of AB, and the circle O passes through three points D, B and C, ∠ doc = 2 ∠ ACD = 90 Verification: the line AC is tangent of circle o

It is proved that: ∵ od = OC, ∵ doc = 90 °,
∴∠ODC=∠OCD=45°．
∵∠DOC=2∠ACD=90°，
∴∠ACD=45°．
∴∠ACD+∠OCD=∠OCA=90°．
∵ point C is on circle o,
The straight line AC is the tangent line of circle o