As shown in the figure, △ ABC, D is on the extension line of BC, and AC = CD, CE is the middle line of △ ACD, and CF bisects ∠ ACB As shown in the figure, △ ABC, D is on the extension line of BC, and AC = CD, CE is the center line of △ ACD, CF bisects ∠ ACB, intersects AB and F, and proves (1) ce ⊥ CF: (2) CF is parallel to AD

As shown in the figure, △ ABC, D is on the extension line of BC, and AC = CD, CE is the middle line of △ ACD, and CF bisects ∠ ACB As shown in the figure, △ ABC, D is on the extension line of BC, and AC = CD, CE is the center line of △ ACD, CF bisects ∠ ACB, intersects AB and F, and proves (1) ce ⊥ CF: (2) CF is parallel to AD

1. Because AC = CD, CE is the midline of △ ACD
So CE ⊥ ad ∠ ace = ∠ DCE
Because CF bisects ∠ ACB, so ∠ ACF = ∠ BCF
Because ∠ ACE + ∠ DCE + ∠ ACF + ∠ BCF = 180 °
So ∠ ACE + ∠ ACF = 90 °
So CE ⊥ CF
2. Because CF ⊥ CE, ad vertical CE
So CF ‖ ad
I wrote it myself!

As shown in the figure, ∠ ACB = 90 ° in △ ABC, CD is high, CE is central line, and ∠ ACD = 3 ∠ 1. (2) 2 degree

[1] Because angle ACB = 90 degrees, angle ACD = 3 angles 1, and angle ACB = angle ACD + angle 1, so angle ACB = 4 angle 1, that is, angle 1 = angle ACB / 4 = 22.5 degrees [2] because the angle ACB of triangle ABC is 90 degrees, CD is high, so triangle ABC is similar to triangle CBD, so angle a = angle 1 = 22.5 degrees, because in right triangle ABC

It is known that, as shown in the figure △ ABC, if the bisector of ∠ ACB intersects AB with E, the bisector of ∠ ACD is CG, eg ‖ BC intersects AC with F, will EF be equal to FG? Why?

EF=FG，
The reason is as follows: ∵ CE bisection ∠ ACB, CG bisection ∠ ACD,
∴∠BCE=∠ECF，∠DCG=∠GCF，
∵EG∥BC，
∴∠FEC=∠BCE，∠G=∠DCG，
∴∠FEC=∠ECF，∠G=∠FCG，
∴EF=CF，CF=FG，
∴EF=FG．

It is known that, as shown in the figure △ ABC, if the bisector of ∠ ACB intersects AB with E, the bisector of ∠ ACD is CG, eg ‖ BC intersects AC with F, will EF be equal to FG? Why?

EF=FG，
The reason is as follows: ∵ CE bisection ∠ ACB, CG bisection ∠ ACD,
∴∠BCE=∠ECF，∠DCG=∠GCF，
∵EG∥BC，
∴∠FEC=∠BCE，∠G=∠DCG，
∴∠FEC=∠ECF，∠G=∠FCG，
∴EF=CF，CF=FG，
∴EF=FG．

It is known that in △ ABC, CE bisects ∠ ACB intersects AB at e, CG bisects exterior angle ∠ ACD. If eg ‖ BD intersects AC at point F, is EF equal to FG? Please state the reasons

EF=FG．
∵ CE bisection ∵ ACB,
∴∠ACE=∠BCE，
∵EG∥BC，
∴∠FEC=∠BCE，
∴∠ACE=∠FEC，
∴EF=FC；
∵ CG bisection ∵ ACD,
∴∠ACG=∠GCD，
∵EG∥BC，∠G=∠GCD，
∴∠G=∠ACG，
∴FG=FC，
∴EF=FG．

It is known that in △ ABC, CE bisects ∠ ACB intersects AB at e, CG bisects exterior angle ∠ ACD. If eg ‖ BD intersects AC at point F, is EF equal to FG? Please state the reasons

EF=FG．
∵ CE bisection ∵ ACB,
∴∠ACE=∠BCE，
∵EG∥BC，
∴∠FEC=∠BCE，
∴∠ACE=∠FEC，
∴EF=FC；
∵ CG bisection ∵ ACD,
∴∠ACG=∠GCD，
∵EG∥BC，∠G=∠GCD，
∴∠G=∠ACG，
∴FG=FC，
∴EF=FG．

As shown in the figure, in RT △ ABC, ∠ ACB = 90 degrees, CD ⊥ AB in D, known as AC = √ 5, BC = 2, then the value of sin ∠ ACD is () A.√5/3 B.2/3 C.2√5/5 D.√5/2

A
You can draw the picture on the draft paper first
∠ ACD = ∠ B (because ∠ B + ∠ BCD = 90 ° and ∠ BCD + ∠ ACD = 90 °)
So sin ∠ ACD = sin ∠ B
According to Pythagorean theorem, ab = 3
So sin ∠ ACD = sin ∠ B = √ 5 / 3, so choose a
Hope to adopt, thank you!

In RT △ ABC, ∠ BAC = 90 ° AB = AC = 2, taking AC as one side, isosceles RT △ ACD are made outside △ ABC, and the length of segment BD is calculated Complete answers are required

Angle ACD = angle BCA = 45 degrees
BC is perpendicular to CD
BC = 2 times root 2, CD = 2 / 2 root sign 2
According to Pythagorean theorem, BD = 2-th root sign 34 can be obtained

Both △ ACD and △ BCE are isosceles right triangles, ∠ ACD = ∠ BCE = 90 °, AE intersects DC at F, BD intersects CE respectively, AE at point G, h.ae is perpendicular to BD?

It is proved that: ∵, ∵ ACD = ∠ BCE = 90 °
∴∠ACE=∠DCB
∵ △ ACD and △ BCE are isosceles right triangles
∴AC=DC EC=BC
∴⊿ACE≌⊿DCB
∴∠CBD=∠CEA
∵∠BCE=90°
∴∠CBD+∠CGB=90°
∵∠CGB=∠EGH
∴∠CEA+∠EGH=90°
∴∠EHG=90°
∴AE⊥BD

As shown in the figure, C is a point on the segment AB, △ ACD and △ BCE are equilateral triangles, AE intersects CD at M BD, CE at point n, and AE at 0 Find the degree of angle AOB Is 2cm equal to CN? Please explain the reason Is 3MN parallel to ab? Please explain the reason

Firstly, SAS is used to explain ⊿ ace ≌ ⊿ DCB
∴∠CAE=∠CDB
∵∠AMC=∠DMO
∴∠AOB=∠CDB＋∠DMO=∠CAE＋∠AMC=180°－60°=120°
2. Not parallel
∵∠ABD＜∠ABE＝∠ACD
3. Parallel
⊿acm≌⊿dcn
∴cm=cn
∵∠mcn=60°
⊿ CMN is an equilateral triangle
∴∠acm=60°=∠cmn
∴mn∥ab