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(1) ∵ D is the midpoint of AB,

As shown in the figure, it is known that D is a point on the extension line of △ ABC side BC, DF is perpendicular to AB and F, AC is at e, ∠ a = 35 ° and ∠ d = 42 ° and the degree of ∠ ACD is calculated I'm sorry, I won't make a picture. I'll write ABC... For you. You can draw it on paper The afbce is connected as a triangle a Fbcde is also a triangle F After connecting, we can see that AFE and ECD are also triangle E I hope you can have a good answer today. I'll add BCD A a is above EC The f afbce is connected in an angle E fbcde is also a triangle After B C D connection, we can see that AFE and ECD are also triangles

In the right triangle AFE, the angle ∠ AEF = 180 ° - ∠ AFE - ∠ a = 180 ° - 90 ° - 35 ° = 55 °
In the triangle CED, ∠ ACD = 180 ° - ∠ CED - ∠ d = 180 ° - 42 ° - 55 ° = 83 ° in the triangle CED

Let d be a point on the extension line of edge BC of triangle ABC, DF is perpendicular to AB and intersects AC and e at F, angle ACD = 86 degrees, angle d = 40 degrees, calculate the degree of angle A

It is shown in Fig
∠2＝∠1＝54°,∠3＝90°
∴∠A＝180°﹣90°﹣54°＝36°

As shown in the figure, in the triangle ABC, AB is equal to AC, point D is the midpoint of BC, De is perpendicular to AC, DF is vertical to AB, and perpendicular feet are respectively E and F. it is proved that de = DF

In △ ABC,
It is known that: ab = AC, ∠ B = ∠ C
It is known that the point D is the midpoint of BC, BD = DC
It is known that De is perpendicular to AC, DF is vertical to AB, and perpendicular feet are e, F,  BFD = ∠ CED = 90 °
In △ BFD and △ CED, ∠ B = ∠ C, BD = DC, ∠ BFD = ∠ CED = 90 °
∴△BFD≌△CED,∴DE=DF.

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．

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?

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∵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．

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．

As shown in the figure, in △ ABC, ∠ C = 90 °, ad is the bisector of ∠ BAC, O is a point on AB, and ⊙ o with OA as radius passes through point D It is proved that BC is ⊙ o tangent

Let AB and ⊙ o intersect point E
∵ ad is the bisector of ∵ BAC,
∴∠BAC=2∠BAD，
And ∵ EOD = 2  EAD,
∴∠EOD=∠BAC，
∴OD∥AC．
∵∠ACB=90°，
Ψ BDO = 90 ° i.e. OD ⊥ BC,
And ∵ od is the radius of ⊙ o,
⊙ BC is the ⊙ o tangent

In the triangle ABC of RT, the angle c = 90 ° AC = 3 BC = 4. The circle is made with C as the center and R as the radius Circle and hypotenuse AB have a common point to find the value range of R

In two cases,
1. Tangent, where the slant height is the radius, 12 / 5 = 2.4
2. It is longer than the short right angle side and shorter than the long right angle side