In the triangle ABC, D is a point on the edge of AC, the angle DBC = angle a, BC = radical 6, AC = 3, find the length of CD Urgent``````

In the triangle ABC, D is a point on the edge of AC, the angle DBC = angle a, BC = radical 6, AC = 3, find the length of CD Urgent``````

∵∠DBC=∠A,∠C=∠C
∴△CBD∽△CAB
∴BC/CD=AC/BC
∴BC²=CD*CA
∵BC=√6,AC =3
∴6=3*CD
∴CD=2

In the triangle ABC, AB is equal to AC equal to 9, angle cab is equal to 120 degrees, ad is the middle line of the triangle, AE is the bisector of angle bad, DF is parallel to the extension line of AB intersection AE At point F, the length of DF is

∵⊥ ABC is an isosceles triangle, D is the midpoint of the bottom,  ad ⊥ BC, ∵∵ BAC = 120 °, bad = 60 °, ADB = 90 °, ∵ AE is the angular bisector of ∠ bad,  DAE = ∠ EAB = 30 °. ∵ DF ∥ AB,  f =  Bae = 30 °. ? DF = DF. ? AB = 9, ∠ B = 3

As shown in Fig. 1, it is known that in RT △ ABC, ∠ ABC = 90 ℃, ∠ cab = 30 ℃, BC = 5. Pass through point a as AE ⊥ AB, and AE = 15, connect be to AC at point P (1) Take point a as the center of the circle and AP as the radius to make ⊙ A. try to judge whether be is tangent to ⊙ a, and explain the reasons; (2) As shown in Fig. 2, under the condition of (1), take the straight line where AB is located as the x-axis and the straight line AE as the y-axis, establish the coordinate system as shown in the figure. Let ⊙ a intersect AB at point Q, and the line L passing through point Q will divide ⊙ a into two parts of 1:3, and find the analytic formula of line L

1. It is known that in RT △ ABC, ∠ ABC = 90?, ∠ cab = 30?, BC = 5. Sina = BC / AC = 1 / 2, so AC = 10, ab = 5 √ 3, AE = 15, AE ⊥ AB, TANE = AB / AE = √ 3 / 3, then E = 30, so AP ⊥ be, then be is tangent to circle A. 2

As shown in the figure, we know that in RT △ ABC, the angle cab = 30 ° BC = 5, passing point a as AE ⊥ AB, and AE = 15, connecting be to point P. crossing point C as CD ⊥ AE, perpendicular foot as D, point a as center r as radius ⊥ a, and point C as radius ⊥ C. if the size of R and R can be changed, and keep ⊥ a tangent to ⊥ C during the change process, and make point d be in the interior of ⊥ a, and point B is outside of ⊥ a, calculate the variation range of R and R except 5

Because ad = 5, ab = 5, root number 3, the variation range of R is 5

As shown in the figure: it is known that in isosceles RT △ ABC, ∠ cab = 90 ° and with ab as the edge, isometric △ abd, AE ⊥ BD, CD and AE intersect at point M. verification: DM = half BC

If we extend Da to the point F, then we have: ∠ CAF = 180 ° - ∠ DAB - ∠ BAC = 180 ° - 60 ° - 90 ° = 30 °. We know that AE is the height of the equilateral △ abd. We can get: de = EB = (1 / 2) BD = (1 / 2) ab. given that ad = AB = AC, we can obtain: ∠ ADC = ∠ ACD = (1 / 2) ∠ CAF = 15 °

As shown in the figure, in RT △ ABC, ∠ C = 90 °, ad bisection ∠ cab, de ⊥ AB in E, if AC = 6, BC = 8, CD = 3 (1) Find the length of de; (2) Find the area of △ ADB

(1) ∵ ad bisection ∵ cab, de ⊥ AB, ∵ C = 90 °,
∴CD=DE，
∵CD=3，
∴DE=3；
(2) In RT △ ABC, AB is obtained by Pythagorean theorem=
AC2+BC2=
62+82=10，
The area of △ ADB is s △ ADB = 1
2AB•DE=1
2×10×3=15．

It is known that in RT △ ABC, ∠ C = 90 ° D is the midpoint of BC edge, de ⊥ AB is e, tanb = 1 2, AE = 7, get De

∵ de ⊥ AB in E,
∴tanB=DE
BE=1
2，
Let de = X,
∴BE=2x，
∴BD=
DE2+BE2=
5x，
∴cosB=BE
BD=2
5，
∵∠C=90°，∴cosB=BC
AB=BE
BD=2
5，
∵ D is the midpoint of BC side, ᙽ BC = 2bd = 2
5x，
∴AB=
Five
2BC=5x，
∵AE=7，
∴AB=AE+BE，
5x=7+2x，
X=7
3．
So de = 7
3．

As shown in the figure, in the triangle ABC, the angle ACB is equal to 90 degrees, CD is perpendicular to ab at point D, AE is bisector angle cab intersects CD at point F, intersection BC is at point E, eg is perpendicular to AB and is perpendicular to point D As shown in the figure, in the triangle ABC, the angle ACB is equal to 90 degrees, CD is perpendicular to ab at point D, AE is bisector angle cab intersects CD at point F, intersection BC is at point E, eg is perpendicular to AB and is perpendicular to point G

∵ AE bisection angle cab AC ⊥ CB eg ⊥ ab
∴AC=AG CD//EG CE=GE
∵∠CAE=∠GAE AF=AF
∴△ACF≌△AGF
∴∠ACF=∠AGF
∵∠ACD+∠DCB=∠DCB+∠CBD=90
∴∠ACD=∠CBD=∠AGF
∴FG//CB
The quadrilateral CFGE is a parallelogram
∵CE=GE
The CFGE is a diamond

As shown in the figure, in the triangle ABC, the angle ACB = 90 degrees. AE bisector angle cab, CD are perpendicular to point D. they intersect point F. is the triangle CFE isosceles triangle? As shown in the figure, in the triangle ABC, the angle ACB = 90 degrees. AE bisector angle cab, CD are perpendicular to point D. they intersect at point F. is the triangle CFE isosceles triangle? Explain the reason

If my picture is correct It should be proved that:
Because AE bisects the angle cab
So angle CAE = angle EAB
Because CD is vertical ab
So the angle fad + angle AFD = 90 degrees
Because angle CAE + angle AEC = 90 degrees
And angle CAE = angle fad
So angle AFD = angle AEC
Because angle AFD = angle CFE
So the angle AEC = the angle CFE
So it's isosceles

As shown in the figure, we know: in the triangle ABC, the angle ACB = 90 degrees, CD is the height of the hypotenuse AB, AE bisection angle cab intersects CD with E, please judge the shape and reason of the triangle efc

It is an isosceles triangle;
Because it is an angular bisector, the angle ead = EAC; because the right triangle, the angle AED = AFC; because the angle AED = CEF, so CE = CF