In the RT triangle ABC, the angle ABC = 90 degrees, the circle O with the diameter of AB intersects AC with D, and the tangent passing through D intersects BC with e. it is proved that de = 1 / 2 BC, If Tan angle c = root 5 divided by 2, de = 2, find ad

In the RT triangle ABC, the angle ABC = 90 degrees, the circle O with the diameter of AB intersects AC with D, and the tangent passing through D intersects BC with e. it is proved that de = 1 / 2 BC, If Tan angle c = root 5 divided by 2, de = 2, find ad

It is proved that the OBE of RT triangle is congruent with ode of RT triangle, so be is= DE.OD=OA=OB It is proved that the angle c is equal to the angle EDC, so de = EC. So de = half of BC
What is tan? You can prove it according to the results of the previous step. I don't understand what you mean by Tan angle C?

As shown in the figure, in RT △ ABC, the angle ACB is 90 degrees. Take BC as the diameter as the center of the circle O, intersect AB with D. e as the midpoint of AC. connect de. prove that De is the tangent of the center o

Connect OE,
Because o and E are the midpoint of two right angle sides of RT △ ABC, RT △ ABC is similar to RT △ EOC,
Therefore, EO / / AB, then ∠ ABC = ∠ EOC, ∠ BDO = ∠ EOD
Because ob = od = radius of the circle, △ OBD is an isosceles triangle, ∠ OBD = ∠ ODB
So, EOC = EOD,
And OC = od = radius of circle, OE is common edge
Therefore, △ EOC ≌ △ EOD is a right triangle, ≌△ EOD, ⊥ Edo = 90 ° de ⊥ OD, that is, ED is perpendicular to the radius of the circle, and D is a point on the circle,
Therefore, De is the tangent of a circle with o as its center and half BC as its radius

As shown in the figure, in RT △ ABC, ∠ ACB = 90 ° CD ⊥ AB, it is proved that BC is the tangent of the circumcircle of △ ADC (2) Which tangent line is the circumscribed circle of △ BDC? Why? (3) If AC = 5, BC = 12, make circle C with C as the center and make circle C tangent to AB, what is the radius of circle C? Urgent and urgent ~ seek the process!

It is proved that: (1) let the circumscribed circle of △ ADC be 0 1
∵ points a, D and C are all on ∵ 1, and ad ⊥ DC
/ / AC is the diameter of 0 1
And ∵ BC ⊥ AC
/ / BC is the tangent of the circumscribed circle of △ ADC
Conclusion
（2）
Let the circumcircle of △ BDC be 0 2
∵ points B, D and C are all on ∵ 2, and BD ⊥ DC
The diameter of BC is 0 2
And ∵ AC ⊥ BC
/ / AC is the tangent line of the circumscribed circle of △ BDC
Conclusion
(3) If point C is taken as the center of the circle so that circle C is tangent to AB, then the radius of the circle must be perpendicular to ab,
So we can only take CD as the radius of the circle
The area of RT △ ABC = AC * BC / 2 = AB * CD / 2
That is, 5 * 12 / 2 = 13 * CD / 2
It can be solved that CD = 60 / 13
I hope my answer will help you

As shown in Figure 1, in the RT triangle ABC, the angle ACB = 90 degrees, AC = 6BC = 8 points, D moves on edge AB, de bisects angle CDB, and intersects edge BC at point E In RT △ ABC, ∠ ACB = 90 °, AC = 6, BC = 8, point d moves on edge AB, de bisects ∠ CDB intersection BC at point E, EM ⊥ BD perpendicular foot is m, en ⊥ CD perpendicular foot is n. (1) when ad = CD, prove: De / / AC; (2) when exploring the value of AD, △ BME is similar to △ CNE? (3) when exploring the value of AD, the area of quadrilateral mend and △ BDE are equal?

(1) (2) 1) 1) when △ BME ⊙ CNE ? 1) when ⊙ 1 ∵ D ∵ D ∵ D ? D ? D ? D ? D ? D ? D ? D ? D ? D ? D ? D ? D ? D \\\\\\\\\\de ∥ AC  be / BC = BD / AB

As shown in the figure, ⊙ o with the diameter of AB of the isosceles ⊥ ABC is crossed with BC at D, and de ⊥ AC at e is made through D. It can be concluded that De is tangent to ⊙ o (1) If point O moves to point B on AB, and the circle with o as its center and ob length as its radius still intersects BC at D, de ⊥ AC, is the above conclusion true? Please state the reasons; (2) If AB = AC = 5cm, Sina = 3 5, then where is the center of circle O at AB, where is ⊙ o tangent to AC?

(1) The reasons are as follows: as shown in the figure, connect OD; ∵ od = ob,  ABC = ∠ ODB, ∵ AB = AC,  ABC = ∠ ACB,  ACB = ∠ ODB, ∵ OD ∵ AC; and ∵ de ⊥ AC, ∵ de ⊥ OD, that is, De is the tangent line of ⊙ O. (2) when o is 3x = 158 from point B on AB, ⊙ o is tangent to ac

As shown in the figure, AB is the diameter of ⊙ o, CD is the chord, CE ⊥ CD intersects AB with E, DF ⊥ CD intersects AB with F, proving that AE = BF

It is proved that if O is used as og ⊥ CD, according to the vertical diameter theorem, og is vertically divided into CD, then CG = DG,
∵CE⊥CD，DF⊥CD，OG⊥CD，
∴CE∥OG∥DF，
∵CG=DG，
∴OE=OF，
∵OA=OB，
∴AE=BF．

As shown in the figure, AB is the diameter of ⊙ o, and CD is the chord. The vertical lines of CD are drawn through two points a and B respectively, and the vertical feet are e and F. verification: EC = DF

It is proved that O is om ⊥ CD is at point M,
∵OM⊥CD，
∴CM=DM，
∵AE⊥EF，OM⊥EF，BF⊥EF，
∴AE∥OM∥BF，
∵ AB is the diameter of ⊙ o,
∴OA=OB，
/ / OM is the median line of trapezoidal aefb,
∴EM=FM
/ / em-cm = fm-dm, i.e., EC = DF

The radius OA and ob of circle O intersect with chord CD at e and f respectively, and CE = cf. it is proved that OE = of; AC = BD

prove:
In the triangle OCF and ode, OC = OD, because CE = DF, so CF = de; because OC = OD, so angle OCD = angle ODC, so triangle OCF and triangle ode are congruent, so of = OE
In triangle OCA and triangle ODB, OC = OD, OA = ob. Because triangle OCF and triangle ode are congruent, so angle COF equals angle doe, so angle COA = angle DOB, so triangle OCA and triangle ODB are congruent, so AC = BD

As shown in the figure, AB is the diameter of ⊙ o, chord de vertically bisects the radius OA, C is the perpendicular foot, chord DF and radius ob intersect with point P, connect EF and EO, if de = double root sign 3, ﹤ DPA = 45 degrees (1) Find the diameter of ⊙ o; (2) Find the area of the shadow part in the graph (PS: This is the 2010 Ningbo high school entrance examination, so please find it yourself.)

1. Connect OD
OC=1/2OD
So the angle doc = 60 degrees
Root 3 = R * sin 60 degrees
R=2
2. Connect of
Angle EDF = DPC = 45 degrees
Angle EOF = 90 degrees
The sector area is 1 / 4Pi * R ^ 2 = Pi
The triangle area is 1 / 2 * 2 * 2 = 2
So the shadow area is PI-2

It is known that: as shown in the graph, △ ABC, ab = AC = 10, D is any point on the edge of BC, respectively make DF ∥ AB intersect AC with F, de ∥ AC intersect AB with E, and calculate the value of de + DF

∵DE∥AC，DF∥AB，
The quadrilateral AEDF is a parallelogram,
∴DE=AF，
And ∵ AB = AC = 10,
∴∠B=∠C，
∵DF∥AB，
∴∠CDF=∠B，
∴∠CDF=∠C，
∴DF=CF，
∴AC=AF+FC=DE+DF=10．
A: the value of de + DF is 10