It is known as follows: in △ ABC, ∠ cab = 120 °, ab = 4, AC = 2, ad ⊥ BC, D are perpendicular feet. Find the length of AD

It is known as follows: in △ ABC, ∠ cab = 120 °, ab = 4, AC = 2, ad ⊥ BC, D are perpendicular feet. Find the length of AD

Crossing point C, do CE ⊥ AB, the extension line of AB and E, ∵, ∵, ∵ cab = 120 °, ab = 4, AC = 2  BC  BC ∵ BC = 120 °, ab = 4, AC = 2  ac * cos angle, cab = 4

As shown in the figure, it is known that in △ ABC, ab = AC, D is a point on AB, de ⊥ BC, e is a perpendicular foot, and the extension line of ED intersects the extension line of Ca at point F, Confirmation: ad = AF

Proof: ∵ AB=AC,
∴∠B=∠C，
∵DE⊥BC，
∴∠C+∠F=90°，∠B+∠BDE=90°，
∵∠ADF=∠BDE，
∴∠F=∠ADF，
∴AD=AF．

In △ ABC, ab = AC, D is a point on AB, and de ⊥ BC at e is made by passing D, and intersects with the extension line of CA to F

∵DE⊥BC
∴∠B+∠BDE=90°,∠C+∠F=90°
∵AB=AC
∴∠B=∠C
FDA = BDE
∴∠F=∠FDA
∴AD=AF

As shown in the figure, in △ ABC, ∠ ACB = 90 °, CE ⊥ AB at point E, ad = AC, AF bisection ∠ cab intersecting CE at point F, and extension line of DF intersecting AC at point G The results showed that: (1) DF ∥ BC; (2) FG = Fe

(1) It is proved that: ∵ AF bisection ∠ cab,
∴∠CAF=∠DAF．
In △ ACF and △ ADF,
A kind of
AC＝AD
∠CAF＝∠DAF
AF＝AF ，
∴△ACF≌△ADF（SAS）．
∴∠ACF=∠ADF．
∵∠ACB=90°，CE⊥AB，
∴∠ACE+∠CAE=90°，∠CAE+∠B=90°，
∴∠ACF=∠B，
∴∠ADF=∠B．
∴DF∥BC．
② Proof: ∵ DF ∥ BC, BC ⊥ AC,
∴FG⊥AC．
∵FE⊥AB，
AF bisection ∠ cab,
∴FG=FE．

As shown in the figure, ad is the angular bisector of △ ABC, the extended line of be ⊥ ad crossing ad is at e, EF ∥ AC is crossing AB to F. it is proved that AF = FB

Proof: ∵ ad bisection ∠ BAC,
∴∠BAD=∠CAD，
∵EF∥AC，
∴∠FEA=∠CAD，
∴∠FAE=∠FEA，
∴FA=FE，
∵BE⊥AD，
∴∠FEA+∠FEB=90°，∠FBE+∠FAE=90°，
∴∠EBF=∠BEF，
∴EF=FB，
∴AF=FB．

As shown in the figure, in the triangle ABC, the angle ACB = 90 degrees, point D is on AB, AC = ad De, vertical CD intersects BC at point e AF, angle BAC intersects BC at point F Verification AF parallel de When AC = 6 AB = 10, find the length of be

prove:
1）
Because: ad = AC, AF bisection ∠ BAC
Therefore: AF is the vertical bisector of the base DC of the isosceles triangle ADC
So: AF ⊥ DC
Because: de ⊥ DC
So: AF / / de
2）
AB=10,AC=6=AD
According to Pythagorean theorem, BC = 8 is obtained
Because: AF bisects DC vertically
So: AF is the median line of the triangle CDE
So: CF = EF = (bc-be) / 2 = (8-be) / 2 = 4-be / 2
Because: De / / AF
So: BD / BA = be / BF = (10-6) / 10 = 2 / 5
So: be = 2 (be + EF) / 5
So: EF = 3bE / 2
Therefore: EF = 4-be / 2 = 3bE / 2
The solution is: be = 2

In the RT triangle ABC, the angle c = 90 °, AC = radical 2, BC = 1, the circle with C as the center and CB as the radius intersects AB at the point P, then the length of AP is obtained

Make the vertical ab of CD, from the area method, we can know that CD = 3 / 3 root sign 6, and from similarity we can get BD = 3 / 3 root sign 3
So BP = 2bd = 2 / 3 root 3, so AP = ab-bp = 3 / 3 root 3

As shown in the figure, we know that in the right triangle ABC, ∠ ABC = 90 ° and ab = ad, CB = CE, try to find the degree of ∠ EBD. (please write the solution process clearly)

Let ∠ a = x °,
∵∠ABC=90°，
∴∠C=（90-x）°，
∵AB=AD，CE=CB，
∴∠ABD=∠ADB，∠BEC=∠EBC，
∴∠ADB=（180−x
2）°=（90-x
2）°，∠EBC=[180-（90-x）]÷2=[45+x
2]°，
∴∠DBC=∠ADB-∠C=（90-x
2）°-（90-x）°=（x
2）°，
∴∠EBD=∠EBC-∠DBC=（45+x
2）°-（x
2）°=45°．

As shown in the figure, we know that in the right triangle ABC, ∠ ABC = 90 ° and ab = ad, CB = CE, try to find the degree of ∠ EBD. (please write the solution process clearly)

Let ∠ a = x °, ∵ ABC = 90 °,  C = (90-x) °, ∵ AB = ad, CE = CB,  abd = ∠ ADB,  BEC = ∠ EBC,  ADB = (180 − x2) ° = (90-x2) °, ∠ EBC = [180 - (90-x)] / 2 = [45 + x2] °, ∵ DBC = ∠ ADB - ∠ C = (90-x2) ° - (90-x) ° = (

As shown in the figure, in RT △ ABC, the circle O with the diameter of right angle side AB intersects the oblique side of D, OE parallels BC and intersects AC with e. it is proved that: (1) De is the tangent of circle o (2) OE is the median line of RT △ ABC

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