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In the isosceles triangle ABC, ∠ a = 120 ° AB = AC = 2cm, the minimum radius of circular paper covering △ ABC is calculated
The minimum circle is that the diameter of the circle is exactly the bottom edge of the isosceles triangle. Because BC = 2 √ 3, the minimum radius = √ 3
As shown in the figure, △ ABC is inscribed in ⊙ o, AB is the diameter of ⊙ o, ﹤ BAC = 2 ∠ B, AC = 6. The tangent line of ⊙ O and the extension line of OC are intersected at point P, and the length of PA is calculated
∵ AB is the diameter of ⊙ o
∴∠ACB=90°
∴∠B+∠BAC=90°
And ∵ BAC = 2 B
∴∠B=30°，∠BAC=60°
∵OA=OC
The △ OAC is an equilateral triangle
∴OA=AC=6，∠AOC=60°
∵ AP is the tangent of ⊙ o
∴∠OAP=90°
In the right angle △ OAP, ∠ P = 90 ° - ∠ AOC = 90 ° - 60 ° = 30 °
∴OP=2OA=2×6=12，
∴PA=
OP2−OA2=
122−62=6
3．
As shown in the figure, △ ABC is the inscribed triangle of ⊙ o, AB is the diameter of ⊙ o, and the point D is on ⊙ o, ﹤ BAC = 35 °, then ∠ ADC=______ Degree
∵ AB is the diameter of ⊙ o,
∴∠ACB=90°；
∴∠B=90°-∠BAC=55°；
According to the circular angle theorem, ∠ ADC = ∠ B = 55 °
In triangle ABC, if angle a minus angle B equals 10 degrees, angle B minus angle c equals 20 degrees, then what is angle a equal to?
A - ∠ B = 10, ① ∠ B - ∠ C = 20, there are ① and ②, we get ∠ A-2 ∠ B + ∠ C = - 10 ③
As shown in the figure, the three vertices of the triangle ABC are on the circle O, and the bisector of the outer corner of the angle ACB intersects the circle O in E and ef ⊥ BD in F The position relation between EO and ab is explored and proved
EO ⊥ bisection AB connects AE and be, because CE is the bisector of ∠ ACD, so: ∠ ace = ∠ ECD, ∠ ECD = ∠ BAE (an outer angle of a circle inscribed with a quadrilateral is equal to the non adjacent inner angle). Therefore, ∠ BAE = ∠ Abe, that is, △ EAB is
As shown in the figure, the three vertices of the triangle ABC are on the circle O, and the bisector of the external angle of ∠ ACB intersects the circle O at E. when the bisector of the outer angle of the triangle ABC intersects with the circle O, EF is perpendicular to BD at f (1) to explore the position relationship between EO and AB, and prove (2) when the shape of triangle ABC changes, whether the value of (BF + CF) △ AC changes? If not, calculate the value, if it changes, ask for the range of change
One
EO ⊥ bisection ab
Connect AE and be
Because CE is the bisector of ∠ ACD, so: ∠ ace = ∠ ECD
And, ∠ ECD = ∠ BAE
So, BAE = ∠ ace
And, ∠ ace = Abe (equal to the circle angle of the arc)
Therefore, ∠ BAE = ∠ Abe
That is, △ EAB is an isosceles triangle
Therefore, EO ⊥ bisects ab
It is proved that the point h. of AB is taken to connect oh
Because △ OAB is an isosceles triangle and H is the midpoint of AB, oh ⊥ ab
Similarly, eh ⊥ ab
So, O is on EH
Then, OE ⊥ bisects ab
Two
On BD, take FG = FC and connect eg
Because of EF ⊥ BD, so, ﹥ CFE = GFE = 90 ⊥
Because CF = GH
EF public
So: RT △ EFC ≌ RT △ EFG
Therefore, FG = CE, and ∠ EGF = ∠ ECF
And, ∠ ECF = ECA (known)
∠ CAE = ∠ CBE (equal to the circle angle of the arc)
AE = be (1 case)
So, △ ace ≌ △ BGF
Therefore, AC = BG = BF + FG = BF + CF
So, (BF + CF) / AC = 1
And this value should not change with the change of △ ABC
The three vertices of the triangle ABC are on the circle O, and the bisector of the outer corner of the angle ACB intersects the circle O on e, EF is perpendicular to BD on f BD is the extension of BC Find the position relation between OE and ab How to prove it needs to be written out No points are given for those copied The number on the second floor is wrong! I've already figured it out by myself. Even AE and be have done EM ⊥ AC certificate.
Because CE is the bisector of angle ACB, the complementary angle of angle ECA is equal to angle ECB, angle ECB is equal to angle EAB, complement angle of angle ECA is equal to angle Abe
The radius of a circle is known to be 4, and a, B, C are the three sides of the inscribed triangle of the circle, if ABC = 16 2, then the area of the triangle is () A. 2 Two B. 8 Two C. Two D. Two Two
∵a
sinA＝b
sinB＝c
sinC=2R=8，
∴sinC=c
8，
∴S△ABC=1
2absinC=1
16abc=1
16×16
2=
2．
Therefore, C
The radius of a circle is known to be 4, and a, B, C are the three sides of the inscribed triangle of the circle, if ABC = 16 2, then the area of the triangle is () A. 2 Two B. 8 Two C. Two D. Two Two
∵a
sinA＝b
sinB＝c
sinC=2R=8，
∴sinC=c
8，
∴S△ABC=1
2absinC=1
16abc=1
16×16
2=
2．
Therefore, C
It is known that the triangle ABC is the inscribed triangle of a circle with radius 4. ABC = 16 pieces of 2, and the area of the triangle is calculated
S = (1 / 2) absinc and: C / sinc = 2R
S=(abc)/(4R)=√2
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