As shown in the figure, the angle B in the right triangle ABC is 90 degrees, the bisector of angle a intersects with point BC at point D, e is a point on AB, de = DC, D is the center of the circle, and D B is the radius of the circle Find that AC is the tangent of circle D

As shown in the figure, the angle B in the right triangle ABC is 90 degrees, the bisector of angle a intersects with point BC at point D, e is a point on AB, de = DC, D is the center of the circle, and D B is the radius of the circle Find that AC is the tangent of circle D

For the two triangles of ADB and ADF, because the angle abd = angle AFD = 90 degrees, angle bad = angle CAD (AD is the bisector of angle a), ad is the common side of the two triangles, so the two triangles are congruent. In this case, DB = DF, because the circle takes DB as the radius, so

Δ ABC is a right triangle, ∠ ABC = 90 ° a circle O with diameter AB intersects AC at point E, point D is the midpoint of BC side, connecting De Tangent to O. De 2. If the radius of circle O is root 3, de = 3, find AE

Connect OE and be
In △ BCE, De is the center line on the hypotenuse, so de = BD = CD,
So angle c = angle CED
In △ AOE, angle a = angle AEO;
Because the angle a + angle c = 90 °,
So the angle AEO + CED = 90 degrees
So the angle OED is 90 degrees
So De is tangent to circle o
In RT △ ABC, ab = 2 times root 3, BC = 2DE = 6
So the angle a is 60 degrees
So △ AEO is an equilateral triangle, AE = radical 3

In the right triangle ABC, the angle c = 90, AC = 3, BC = 4. The circle with point C as the center and Ca as the radius intersects with point d respectively, and the lengths of AB and AD are calculated

Should be in D, e two points. Find the length of AE and AD
Suppose D is on AB and E is on BC
(1) Seek AE
Then AE = √ (AC ^ 2 + CE ^ 2) = √ (3 ^ 2 + 3 ^ 2) = 3 √ 2
(2) Seek ad
As DF ⊥ AC, the intersection point is f, and AF = 3x
Then DF = (4 / 3) * 3x = 4x
Pythagorean theorem: ad ^ = DF ^ 2 + AF ^ 2, ad = 5x
(3-3X)^2+(4X)^2=3^2
X=18/25
Therefore, ad = 18 / 5

As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, AC = BC, D is the midpoint of BC edge, CE ⊥ ad intersects point E, BF ∥ AC, and the extension of CE is at point F Confirmation: BD = BF

It is proved that: in ∵ RT △ ABC,

In the isosceles right triangle ABC, the angle ACB = 90 ', D is the midpoint of BC. CE is perpendicular to e, BF / / AC, CE extends to point F. verification: ab vertically bisects DF

From the meaning of the title
∠CAD=∠FCB,AC=BC,∠ACD=∠CBF=90
So △ ACD ≌ △ CBF
CD=BD=BF
Because △ BDF is an isosceles right triangle
So ∠ FDB = 45
Because CBA = 45
So ab ⊥ DF
Because BD = BF
So AB bisects DF vertically

In the right triangle ABC, the angle ACB is equal to 90 degrees, AC is equal to BC, point D is the midpoint of BC, CE is perpendicular to ad perpendicular e, BF is parallel to AC, the extension of CE is at point F, Prove that AC is equal to 2BF

∵CE⊥AD AC⊥BC
∴∠BCF=90°-∠ADC=∠CAD
∵BF∥AC
⊥ BC and AC = CB
∴△CBF≌△ACD
∴BF=CD=BC/2=CA/2
∴AC=2BF

Δ ABC is an isosceles right triangle, CF ⊥ ad, D is the midpoint of BC. Find BF / ad

Let ∠ ABC = 90 °
Let BD = 1, then AB = BC = 2, ad = √ 5 BF = 1 × 2 / √ 5 BF / ad = 2 / 5 = 0.4
[if ∠ C = 90 ° C, the calculation of BF is more troublesome. For ⊿ BDF, the cosine theorem can be used

As shown in the figure, △ ABC is a right triangle, ∠ ACB = 90 °, CD ⊥ AB is at D, e is the midpoint of AC, and the extension of ED and CB intersect at point F (1) The results showed that fd2 = FB · FC; (2) If G is the midpoint of BC, connecting Gd, is GD perpendicular to ef? And explain the reason

(1) It is proved that ∵ e is the midpoint of the slanted edge of RT △ ACD,
∴DE=EA，
Ψ a = ∠ 2, (1 point)
∵∠1=∠2，
Ψ 1 = ∠ a, (2 points)
∵∠FDC=∠CDB+∠1=90°+∠1，∠FBD=∠ACB+∠A=90°+∠A，
∴∠FDC=∠FBD，
∵ f is the common angle,
﹤ FBD ∽ FDC. (4 points)
∴FB
FD＝FD
FC．
/ / fd2 = FB · FC. (6 points)
(2) GD ⊥ EF. (7 points)
The reasons are as follows.
∵ DG is the center line on the inclined side of RT △ CDB,
∴DG=GC．
∴∠3=∠4．
From (1) ∵ ∵ FBD ∵ FDC,
∴∠4=∠1，
Ψ 3 = ∠ 1. (9 points)
∵∠3+∠5=90°，
∴∠5+∠1=90°．
⊥ EF. (10 points)

The triangle ABC is a right triangle ∠ ACB = 90 ° and CD ⊥ AB at De is the extension of the midpoint of AC, and the extension line of ED and CB intersect at F And (2) if G is the midpoint of BC connecting Gd, is GD perpendicular to ef? And explain the reasons

What is the first question?
FD 2 = f? Can you make it clear?
Second, vertical, because de and DG are the midlines on the hypotenuse of a right triangle, EC = ed, DG = CG
So the angle EDC = ECD GCD = GDC
EDC+GDC=ECD+GCD=90

The triangle ABC is a right triangle angle ACB = 90 degrees, CD is perpendicular to d 20, AC midpoint ED is the extension line at F.1, find fd2

That's what you're asking?
A right triangle ABC, angle c = 90 degrees, CD perpendicular to AB, e is the midpoint of AC, ed extension line and CB intersect F, it is proved that FD ^ 2 = FB * FC
To prove that FD ^ 2 = FB * FC, that is, to prove FD / FB = FC / FD, that is to say, to prove △ FDB ∽ △ FCD
In RT △ ADC, the center line on the hypotenuse is equal to half of the hypotenuse, so de = CE, ∠ DCE = ∠ CDE
Because ∠ ade + ∠ CDE = ∠ FCD + ∠ DCE = 90 °, so ∠ ade = ∠ FCD
And ∠ ade and ∠ FDB are antiparietal angles, so ∠ FDB = ∠ FCD
So △ FDB ∽ △ FCD, FD / FB = FC / FD