As shown in the figure, in △ ABC, ∠ ABC = 2 ∠ C, ad is the height on BC edge, extend AB to point E, make be = BD, pass through point D, e lead straight line to AC at point F, then AF = FC, why?

∵BE=BD
∴∠E=∠BDE
∴∠ABC=∠E+∠BDE=2∠E
∴∠C=∠E=∠BDE
And ∠ BDE = ∠ FDC
∴∠FDC=∠C
∴FD=FC
∵ ad is high
∴∠ADF+∠FDC=90°
And ∠ C + ∠ DAC = 90 ° and ∠ FDC = ∠ C,
∴∠ADF=∠DAC，
∴AF=FD
∴AF=FC．

In the triangle ABC, the angle c is equal to 90 degrees, AC is equal to 3, BC is equal to 4, and point D is on ab. take ad as the diameter, make a circle O, cut BC to e, and find the radius of circle o

The common angle is CBA
The results show that △ ABC is similar to △ OEB;
∴OE/AC=OB/AB
∵ ABC is a right triangle
∴AB^2=AC^2+BC^2=25'
AB=5
Let the radius of circle o be r
OE=R
OB=AB-R
therefore
R/3=(5-R)/5;
R=1+7/8

In RT △ ABC, ∠ C = 90 °, AC = 3, BC = 4. What is the position relationship between a circle with C as its center and R as its radius and ab? Why? （1）r=2；（2）r=2.4；（3）r=3．

Make CD ⊥ AB in D,
In the right triangle ABC, according to the Pythagorean theorem, ab = 5, then
CD=AC•BC
AB=2.4；
(1) When r = 2, 2.4 > 2, the line and the circle are separated;
(2) When r = 2.4, the line is tangent to the circle;
(3) When r = 3, 2.4 < 3, the line intersects the circle

In the RT triangle ABC, the angle c is equal to 90 degrees. Take BC as the diameter to make a circle O, intersect AB with point D, take AC point E, connect de and OE. Find that De is the tangent of circle o

prove:
Because e is the midpoint of AC, CO = Bo
So OE is the median of △ ABC,
So OE ‖ AB,
So ∠ COE = ∠ B, ∠ EOD = ∠ ODB,
Od = ob,
So ∠ ODB = ∠ B,
So ∠ EOC = ∠ EOD,
Co = do, EO is the common edge
So △ EOC ≌ △ EOD
So ∠ ADO = ∠ ACO,
Because the angle c is 90 degrees,
So ∠ Edo = 90 °
So De is the tangent of circle o

As shown in the figure, in the triangle ABC, ab = AC, ∠ BAC = 120 ° to find the value of AB: BC 1. In the triangle ABC, ab = AC, ∠ BAC = 120 ° to find the value of AB: BC 2 the ratio of the two line segments is 3:5, and their difference is 5cm. So the sum of these two lines is___________ .

In the right triangle abd, BD = (√ 3) / 2) AB, then BC = √ 3AB

As shown in the figure, in △ ABC, ab = AC, ⊙ o with ab as diameter intersects BC at point m and Mn ⊥ AC at point n (1) It is proved that Mn is the tangent of ⊙ o; (2) If ∠ BAC = 120 ° AB = 2, calculate the area of shadow part in the graph

(1) It is proved that: connect OM. ∵ om = ob,  B = ∠ OMB.

As shown in the figure, given the triangle ABC, ab = AC, angle BAC = 120 degrees, calculate the value of AB ratio BC It's an isosceles triangle There is no picture

1: Radical 3

As shown in the figure, △ ABC is inscribed in ⊙ o, ∠ BAC = 120 ° and ab = AC = 4. Find the diameter of ⊙ o

Connect Bo and extend the intersection circle O to point D and connect ad,
∵∠BAC=120°，AB=AC=4，
∴∠C=30°，
∴∠BOA=60°．
And ∵ OA = ob,
The △ AOB is an equilateral triangle
∴OB=AB=4，
∴BD=8．
The diameter of ⊙ o is 8

Given that the isosceles triangle ABC (AB = AC) is inscribed in the circle O, ∠ BOC = 130 ° then ∠ A is equal to

Because in △ BCO, ∠ BOC = 130 ° so ∠ OBC + ∠ OCB = 180 ° - 130 ° = 50, and because point O is the inscribed circle of △ ABC, Bo and Co are angular bisectors of ∠ ABC and ∠ ACB, i.e., ∠ ABC + ∠ ACB = 2 (∠ OBC + ∠ OCB) = 100 °, so ∠ a = 180 ° - (∠ ABC + ∠ ACB) = 180 ° - 100 °

The isosceles triangle ABC, ab = AC, angle BAC = 120 degrees, P is the midpoint of BC The isosceles triangle ABC, ab = AC, angle BAC = 120 degrees, P is the midpoint of BC. Xiaohui holds a transparent triangular plate with 30 degree angle, so that the vertex of 30 degree falls on point P, and the triangle plate rotates around point P. it is shown that the triangle BPE is similar to the triangle CEP.)

When point E is rotated to the extension line of AP (i.e. when a, P and E are on the same line), triangle BPE is similar to triangle CEP

As shown in the figure, in △ ABC, ∠ ABC = 2 ∠ C, ad is the height on BC edge, extend AB to point E, make be = BD, pass through point D, e lead straight line to AC at point F, then AF = FC, why?