In the triangle ABC, (isosceles triangle), ab = AC, angle BAC = 120 degrees, ad is perpendicular to AC, if the square of AB = 3, then BC The figure is a large isosceles triangle. The large triangle has a small isosceles triangle. A is the vertex. D is a point on BC

Make de perpendicular to ab. e is the vertical point (on AB)
Because ABD is isosceles triangle,
So AE = be = 2 / 2 root 3
Because the angle abd = 30 degrees
So according to the cosine theorem, BD = ad = 1
Because the angle of ACD is 1 degree
So according to the sine theorem CD = 2
So BC = BD + CD = 3

In the obtuse triangle ABC, if AB = AC, D is a point on BC and ad divides △ ABC into two isosceles triangles, then the degree of ∠ BAC is () A. 150° B. 124° C. 120° D. 108°

Let ∠ ABC be X
（180°-x）÷2+x+2x=180°
X = 36 ° is obtained
∴180°-36°×2=108°．
Therefore, D

As shown in the figure, the point a B C D is on the circle O, and the angle ADC = angle EDB = 60 degrees. It is proved that the triangle ABC is an equilateral triangle No picture, sorry, if you can do it yourself, sorry

Point E is the intersection of AB and CD
From ∠ ADC = 60 ° and ∠ ADC = ∠ ABC,
∴∠ABC=60°,
Similarly: ∠ EDB = 60 ° and ∠ EDB = ∠ BAC,
∴∠BAC=60°,
The △ ABC is an equilateral triangle

As shown in the figure, in △ ABC, ab = AC, ⊙ o with ab as the diameter intersects AC and E, and crosses BC and D (1) D is the midpoint of BC; （2）△BEC∽△ADC；（3）BC2=2AB•CE．

It is proved that: (1) ∵ AB is the diameter of ⊙ o,
∴∠ADB=90°，
That is, ad is the height on the bottom BC
And ∵ AB = AC,
ν Δ ABC is an isosceles triangle,
ν D is the midpoint of BC;
(2) ∵ CBE and CAD are
The circumferential angle of De,
∴∠CBE=∠CAD，
And ∵ BCE = ∵ ACD,
∴△BEC∽△ADC；
(3) From △ BEC to △ ADC, we can know CD
CE=AC
BC，
That is, CD · BC = AC · CE,
∵ D is the midpoint of BC,
∴CD=1
2BC，
And ∵ AB = AC,
∴CD•BC=AC•CE=1
2BC•BC=AB•CE，
BC2 = 2Ab · CE

As shown in the figure, the three vertices of the equilateral triangle ABC are all on ⊙ o, and D is At any point of AC (not coincident with a and C), then the degree of ∠ ADC is______ Degree

0

(BF + CF) / AC will not change
The content of this theorem is Archimedes' broken string theorem, which is similar to this one
You just connect be and CE and do EM perpendicular to point M
Then prove that △ AEM and △ bef are congruent
This leads to the conclusion that AF = BF
So BF + CF = am + cm
So (BF + CF) / AC = 1, it remains unchanged

As shown in the figure, △ ABC, AC = BC, ⊙ o with BC as diameter intersects AB at point E, and line EF ⊥ AC at F. verification: EF is tangent to ⊙ o

Proof: connect OE, CE,
∵ BC is the diameter of circle o,
∴∠BEC=90°，
⊥ AB, and AC = BC,
Ψ e is the midpoint of AB, and O is the midpoint of diameter BC,
The OE is the median line of △ ABC,
∴OE∥AC，
∴∠AFE=∠OEF，
EF ⊥ AC,  AFE = 90 °,
∴∠OEF=90°，
Then EF is the tangent of circle o

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 area of inscribed triangle with radius 1 is 0.25,

0

It's an indefinite value with infinite solutions
You can solve this problem by adding:
∵ a triangle inscribed in a circle is a regular triangle
The center of the triangle is the center of the circumscribed circle, that is, the distance from the center to each vertex is equal to the radius of the circumscribed circle
The length from the center to the vertex of an equilateral triangle is equal to = 20
This center is also the center of gravity of an equilateral triangle. The distance from the center of gravity to one side is half the length to the vertex, which is equal to 10
The height of the total equilateral triangle is: 20 + 10 = 30
Therefore, the side length of an equilateral triangle is 2 / 3 √ 3 × 30 = 20 √ 3
Area of triangle =1/2 × 20 × 3 × 30=300 × 3

In the triangle ABC, (isosceles triangle), ab = AC, angle BAC = 120 degrees, ad is perpendicular to AC, if the square of AB = 3, then BC The figure is a large isosceles triangle. The large triangle has a small isosceles triangle. A is the vertex. D is a point on BC