It is known that in the RT triangle ABC, ∠ ACB = 90 ° CD is perpendicular to AB and D, and AF bisects ∠ cab As shown in the figure, in the RT triangle ABC, it is called ACB = 90 degrees, CD is vertical AB, vertical foot is D, AF bisection is called cab in E, intersection CB in F, and FG parallel AB intersects CB in G. it is proved that CF = GB

It is known that in the RT triangle ABC, ∠ ACB = 90 ° CD is perpendicular to AB and D, and AF bisects ∠ cab As shown in the figure, in the RT triangle ABC, it is called ACB = 90 degrees, CD is vertical AB, vertical foot is D, AF bisection is called cab in E, intersection CB in F, and FG parallel AB intersects CB in G. it is proved that CF = GB

The paragraph "AF bisection is called cab in E, crossing CB in F" should be changed to: AF bisection cab, cross CD to e, and BC to F
If FM ⊥ AB is set at m through F, then FM ∥ CD  BFM = ∠ GCD, ∠ BMF = ∠ GIC = 90 degrees
∵ CD vertical AB, vertical foot D, ∵ ACB = 90 degrees
Ψ AED + ∠ BAF = 90 ∠ CAF + ∠ AFC = 90 degrees
∵∠CAF=∠BAF
∴∠AED=∠CEF=∠AFC
∴CF=CE
And ∵ CAF = ∵ BAF FC ⊥ AC FM ⊥ ab
∴FC=FM
∴△CEG≌△FMB
∴CG=BF
∴CG-FG=FB-FG
That is, CF = GB

As shown in the figure, in the RT triangle ABC, the angle ACB = 90 degrees, AC = BC = 6

In order to make the quadrangle qpcp 'be rhombic, it is necessary to PC = PQ (ac-ad) 2 + PD? 2 = PE? + (bc-ec-bq) ∵ AP = √ 2T,  ad = PD = EC = t (6-T) 2 + T? = (6-T) 2 + (6-t-t) 2, which can be reduced to: t? - 8t + 12 = 0 (T-2) (T-6) = 0 ﹤ T1 = 2, T2 = 6 (round off

(2000, Henan) as shown in the figure, in isosceles RT △ ABC, ∠ C = 90 ° D is any point of the hypotenuse AB, AE ⊥ CD is at e, BF ⊥ CD is the extension of CD at F, CH ⊥ AB is at h, and AE is at g. it is proved that BD = CG

It is proved that ∵ ABC is an isosceles right triangle, CH ⊥ ab,
∴AC=BC，∠ACH=∠CBA=45°．
∵CH⊥AB，AE⊥CF，
∴∠EDH+∠HGE=180°．
∵∠AGC=∠HGE，∠HDE+∠CDB=180°，
∴∠AGC=∠CDB．
In △ AGC and △ CDB,
∠ACG＝∠CBD
∠AGC＝∠CDB
AC＝CB ，
∴△AGC≌△CDB（AAS）．
∴BD=CG．

In the RT triangle ABC, the angle ACB = 90 degrees, the angle a = 30 degrees, BC = 1, AC =?

Angle a = 30 degrees, BC = 1, then AB = 2BC = 2, then AC ^ 2 = AB ^ 2-bc ^ 2 = 2 ^ 2-1 ^ 2 = 3
So AC = radical 3

RT triangle ABC, angle ACB = 90 degrees, D is the midpoint of BC, CE is perpendicular to AD and E, and the angle DBE = angle DAB

From the projective theorem, CD ^ 2 = de * Da
Because CD = DB
So DB ^ 2 = be * Da
So the triangle DBE is similar to the triangle DAB
So angle DBE = angle DAB

In RT △ ABC, ∠ ACB = 90 ° CE vertical AB, ad bisect ∠ cab intersecting CE and point F, and FG ‖ AB intersecting CB and G, then the relationship between Cd and BG is As above 111111C 1111111111D 111111F11111 G A 1111E1111111111B 1 is a blank letter, and the position is shown in the figure

Tips:
Make DM ⊥ AB in M
Prove CE = CD (∠ AED = ∠ CDE)
Prove again △ CEG ≌ △ DMB (AAS)
CG = BD is obtained
∴CD=BG

In RT △ ABC, the angle c = 90. AC = 4. BC = 2 AC.BC If you circle the diameter, the area of the shadow is

Is the shadow part two crescent shaped?
Area of shadow part = area of triangle = 4 × 2 △ 2 = 4

As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 4, BC = 2, draw semicircles with AC and BC as diameters respectively, then the area of shadow part in the figure is______ (the result retains π)

Let the area of each part be: S1, S2, S3, S4, S5, as shown in the figure, ∵ the sum of the areas of the two semicircles is: S1 + S5 + S4 + S2 + S3 + S4, △ ABC's area is S3 + S4 + S5, the area of shadow part is S1 + S2 + S4,

As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 4, BC = 2, draw semicircles with AC and BC as diameters respectively, then the area of shadow part in the figure is______ (the result retains π)

Let the area of each part be: S1, S2, S3, S4, S5, as shown in the figure, ∵ the sum of the areas of the two semicircles is: S1 + S5 + S4 + S2 + S3 + S4, △ ABC's area is S3 + S4 + S5, the area of shadow part is S1 + S2 + S4,

As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 4, BC = 2, draw semicircles with AC and BC as diameters respectively, then the area of shadow part in the figure is______ (the result retains π)

Let the area of each part be: S1, S2, S3, S4, S5, as shown in the figure, ∵ the sum of the areas of the two semicircles is: S1 + S5 + S4 + S2 + S3 + S4, △ ABC's area is S3 + S4 + S5, the area of shadow part is S1 + S2 + S4,