Ad, AE and CB are all tangent lines, and the tangent points are D, e, F, ad = 4, so we can find the circumference of the triangle ABC

Ad, AE and CB are all tangent lines, and the tangent points are D, e, F, ad = 4, so we can find the circumference of the triangle ABC

Either give the diagram or write the title in detail

It is known that the triangle ABC is inscribed in the circle O, the point D is on the long line of OC, SINB = 1 / 2, and the angle d = 30 degrees 1. 2. If AC = 6, find the length of AD

Because SINB = 1 / 2, the angle B = 30 degrees, the angle AOC = 60 degrees (the center angle is twice the circumference angle), and the point D is on the long line of OC, and the angle d = 30 degrees
Therefore, in the triangle oad, the angle oad = 90 degrees, that is, ad is the tangent of circle o
At the same time, the center angle of the circle AOC=60 degrees, OA=OC (equal radius), that is, the triangle AOC is an equilateral triangle, so AO=AC=6
Because the triangle oad is a right triangle with angle d = 30, OD = 2 * Ao = 2 * 6 = 12
According to the Pythagorean theorem, the square of ad = the square of OD - the square of OA = 12 * 12-6 * 6 = 108
Ad = 6 * radical 3 = 10.392 (approximately equal to)

It is known that the inscribed circle O, D of the triangle ABC is on the extension line of OC, the angle B = 30 degrees and the angle CAD = 30 degrees 1: Proof: ad is tangent of circle o 2: If CD is perpendicular to AB, BC = 5, find the length of AD Using the method of grade three

1. It is proved that the arc corresponding to the extended co intersecting circle O to e, connecting AE, OA ? B and  e are all inferior arcs AC,  B = 30 ? OA = OE  OAE  e = 30 ? diameter CE ? CAE = 90 ? OAC ? OAE = 60 M CAD = 30 ? CAD = 30

The triangle ABC is inscribed in the circle O, the angle bad = angle CAD, de parallel AB, de intersect AC point P, intersect circle O at e. it is proved that OD is the perpendicular line of BC The triangle ABC is an arbitrary triangle, O is the center of the circle, ad is the bisector of the angle BAC, intersecting the circle at point D, connecting OD, making De, parallel AB, De, AC, crossing point P and connecting op. (please draw your own picture) (1) It is proved that OD is the perpendicular of BC; (2) It is proved that AC = De; (3) Prove the Po bisection angle APD

(1) Connecting OB and OC, ad is the bisector of the angle BAC. According to the arc, we can get that BD arc = CD arc, angle BOD = angle codbo = CO (radius), the intersection point of OD and BC is m, the edges and corners are equal, and the 3-angle BOM is equal. The angle OMC = angle OMB, BM = cm, angle BMC = 180 degrees = angle OMC + OMB, and the angle OMC = 90 degrees is obtained

As shown in the figure, we know that the circumference of △ ABC is 21, ob, OC bisect ∠ ABC and ∠ ACB respectively, OD ⊥ BC is in D, and OD = 3, and the area of ⊥ ABC is______ .

Make OE ⊥ AC, of ⊥ AB, the vertical feet are e and F, connect OA, ∵ ob, OC, bisect ⊥ ABC and ⊥ ACB, OD ⊥ BC, ⊥ od = OE = of, ⊥ ABC = s ⊥ OBC + s ⊥ OAB = 12 × OD × BC + 12 × OE × AC + 12 × of × AB = 12 × OD × (BC + AC + AB) = 12 × 3 × 21 = 31.5

It is known that in the triangle ABC, BD is perpendicular to AC and D, CE is perpendicular to AB and point E, and points m and N are the midpoint of BC and de respectively

Connect me and MD to form △ Med
∵ △ EBC and △ DBC are right triangle
And M is the midpoint on the hypotenuse of two right triangles
∴ ME=MD=(1/2)BC
Therefore, △ Med is an isosceles triangle
And N is the midpoint of the bottom edge of the triangle,
Therefore, Mn ⊥ de

As shown in the figure, in the triangle ABC, the angle ACB = 90 degrees, AC = BC, the straight line Mn passes through point C, and ad is perpendicular to point D, and be is perpendicular to point E

[1] When the line Mn is rotated to the position of Figure 1 around point C, it is proved that de = AD + be; \ x0d [2]. When the line Mn rotates around point C to the position in Figure 2, does the conclusion in [1] still hold? If yes, please give proof; if not, please explain the reasons

As shown in the figure, in the triangle ABC, the angle is ABC=90 °, AC=BC. the straight line MN passes through point C and AD ⊥ MN is in D, BE ⊥ MN is in E. Ask what is the relationship between DE, AD and BE

It is impossible for a right angle edge to be equal to an oblique edge
If it is changed to ∠ ACB = 90 degrees, then there is ad = be + De
It can be proved that triangle ADC and triangle BEC are congruent

It is known that in △ ABC, ∠ ACB = 90 ° AC = BC, the straight line Mn passes through point C, and ad ⊥ Mn is in D, be ⊥ Mn is in E Results: ① △ ADC ≌ △ CEB; ② de = ad-be

It is proved that: ① ∵ ACB = 90 °, be ⊥ CE, ad ⊥ CE,
∴∠BEC=∠ACB=∠ADC=90°，
∴∠ACE+∠BCE=90°，∠BCE+∠CBE=90°，
∴∠ACD=∠CBE，
In △ ADC and △ CEB
∠ADC＝∠BEC
∠ACD＝∠CBE
AC＝BC ，
∴△ADC≌△CEB（AAS）．
②∵△ADC≌△CEB，
∴AD=CE，BE=CD，
∴CE-CD=AD-BE，
∵DE=CE-CD，
∴DE=AD-BE．

It is known that the circle O is the circumscribed circle of △ ABC, and ab is the diameter. If PA is perpendicular to AB, (P is outside the circle), Po passes through the midpoint m of AC. it is proved that PC is tangent of circle o

Because triangle PAB is equal to triangle PCB, PC is tangent of circle o