![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
As shown in the figure, △ ABC is an isosceles triangle, ∠ ACB = 90 ° and the midpoint D of BC is de ⊥ AB, and the perpendicular foot is e to connect CE. Find the value of sina ∠ ace
Suppose be = 1, then de = 1, BD = √ 2, BC = 2 √ 2 = AC, ab = 4, AE = 3, EF = 3 √ 2 / 2. CF = √ 2 / 2
As shown in the figure, the triangle ABC is an isosceles triangle, and the angle ACB is 90 degrees. De is made through the midpoint D of BC, and is perpendicular to ab. connect CE to find the value of sin angle ace
Let de = 1, then because ABC is an isosceles right triangle, be = 1, CD = DB = root 2, AC = 2, root 2, AE = 3, so CE ^ 2 = AE ^ 2 + AC ^ 2-2ac × AE × cos angle a, so CE = radical 5, CE / Sina = AE / sin angle ace, so sin angle ace = (3 root sign 10) / 10
As shown in the figure, △ ABC, D and E are the midpoint of edges BC and ab respectively, and AD and CE intersect at G Verification: GE CE＝GD AD＝1 3．
Proof: connected to ed
∵ D and E are the midpoint of side BC and ab respectively,
∴DE∥AC，DE
AC＝1
2，
∴∠ACG=∠DEG，∠GAC=∠GDE，
∴△ACG∽△DEG．
∴GE
GC＝GD
AG＝DE
AC＝1
2，
∴GE
GE+CG=GD
GD+AG，
∴GE
CE＝GD
AD＝1
3．
As shown in the figure, the triangles ABC, D and E are the midpoint of BC and ab respectively
I've just done it. There's a picture. There's no need to post it. Wait a minute
Connecting de
Because D and E are the midpoint of sides BC and ab respectively
So De is the median line
So de ‖ AC and de = AC / 2
So △ DEG ∽ ACG
So CG / Ge = Ag / Gd = AC / de = 2
Therefore, the ratio of Ge / 1 + 1 + 1 + 1 + 1
So (GE + CG) / Ge = (GD + Ag) / Gd = 3
CE / Ge = ad / Gd = 3
So Ge / CE = GD / ad = 1 / 3
What do not understand can continue to ask, at any time online
In the triangle ABC, D and E are the midpoint of sides BC and ab respectively. AD and CE intersect at point g. it is proved that GE divided by CE is equal to GD divided by AD
DE//1/2AC
EG=1/2GC EG=1/3EC EG/EC=1/3
DG=1/2GB DG=1/3AD DG/AD=1/3
As shown in the figure, in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at D, e is a point on AB, AF ⊥ CE is at F, ad intersects CE at G point. It is proved that ∠ B = ∠ CFD
It is proved that in RT △ AEC, AF ⊥ EC,
∴AC2=CF•CE．
In RT △ ABC, ad ⊥ BC,
∴AC2=CD•CB．
∴CF•CE=CD•CB．
∴CF
CB＝ CD
CE．
∵∠DCF=∠ECB，
∴△DCF∽△ECB．
∴∠B=∠CFD．
As shown in the figure, in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at D, e is a point on AB, AF ⊥ CE is at F, ad intersects CE at G point. It is proved that ∠ B = ∠ CFD
It is proved that in RT △ AEC, AF ⊥ EC,
∴AC2=CF•CE．
In RT △ ABC, ad ⊥ BC,
∴AC2=CD•CB．
∴CF•CE=CD•CB．
∴CF
CB＝ CD
CE．
∵∠DCF=∠ECB，
∴△DCF∽△ECB．
∴∠B=∠CFD．
As shown in the figure, in △ ABC, D is the midpoint of BC, the straight line GF passing through point d intersects AC with F, parallel line BG crossing AC with point G, de ⊥ GF, crossing AB at point E, connecting eg (1) Results: BG = CF; (2) Please judge the relationship between be + CF and EF and prove your conclusion
It is proved that: (1) ∵ BG ∥ AC,
∴∠DBG=∠DCF．
∵ D is the midpoint of BC,
∴BD=CD
And ∵ BDG = ∵ CDF,
In △ bgd and △ CFD,
A kind of
∠DBG＝∠DCF
BD＝CD
∠BDG＝∠CDF
∴△BGD≌△CFD（ASA）．
∴BG=CF．
（2）BE+CF＞EF．
∵△BGD≌△CFD，
∴GD=FD，BG=CF．
And ∵ de ⊥ FG,
/ / EG = ef (the distance from the vertical bisector to the end of the line segment is equal)
In △ EBG, be + BG > eg,
Be + CF > EF
As shown in the figure, in RT △ ABC, ∠ BAC = 90 ° ad ⊥ BC bisects ∠ ABC, EF ∥ BC in D, BG and intersects AC
Verification: AE = CF: AE = CF, pass e point as the parallel line of AC, intersection AB at P, intersection BC at Q ? BAC = 90 °, and PQ ? AC EPB = 90 PAE + ∵ pea = 90 ∵ ∵ ad ? BC DEQ eqd = 90 ⊙ 8857; ad ? BC 9 DEQ ᙽ pea = 90 △ BEQ ν AE =
In △ ABC, the bisector of ∠ B intersects with the bisector of the outer corner of ∠ C at point D, DG ‖ BC, AC, AB at f and g. it is proved that GF = bg-cf P.24
∵, the bisector of ∵ B intersects with the bisector of C at point D, DG ∥ BC, AC, AB at f and G
The quadrilateral bcdg is a parallelogram
It can be proved that ∠ FCD = ∠ FPC, FC = FD
Similarly, it can be proved that ∠ GBD = ∠ GDB, GB = GD
∵GF=GD-FD
∴GF=BG-FC
Come on
RELATED INFORMATIONS
- 1. In the triangle ABC, the bisector of angle B and the bisector of angle c intersect at point DDG / BC, and acab is proved to be GF = bg-cf at two points of FG
- 2. As shown in the figure: in isosceles RT △ ABC, ∠ C = 90 degrees, AC = 8, f is the midpoint of AB edge, points D and e move on AC and BC sides respectively, and keep ad = CE, connect We need the whole proof process. Thank you
- 3. [function] point P is an image of inverse scale function and positive scale function y = - 2x. PQ is perpendicular to X axis, and the coordinate of perpendicular foot is (2,0) If point m is on the inverse scale function image and the area of △ MPQ is 6, find the coordinates of point M Missing two words Point P is the intersection of an image with an inverse scale function and a positive scale function y = - 2x
- 4. The graph of the known quadratic function y = AX2 + BX + C (a ≠ 0) is shown in the figure ① ABC > 0; ② B < A + C; ③ 4A + 2B + C > 0; ④ 2C < 3B; ⑤ a + b > m (am + b) (m ≠ 1 real number) The correct conclusion is () A. 2 B. Three C. Four D. Five
- 5. As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained
- 6. D is the middle of the edge BC of the triangle ABC, De is vertical AC, DF is vertical AB, the vertical feet are e, F, and BF = CE, judge the shape of triangle ABC Is it an isosceles right triangle? Give proof
- 7. As shown in the figure, in RT △ ABC, ab = AC, ∠ a = 90 °, point D is any point of BC, DF ⊥ AB is in F, de ⊥ AC is in E, M is the midpoint of BC. Try to judge what shape △ MEF is and prove your conclusion
- 8. As shown in the figure, in the isosceles triangle ABC, ab = AC, D is a point on the extension line of AB, BD = AB, CE is the center line on waist AB, and verify; CD = 2ce needs specific process
- 9. In △ ABC, ab = BC, ∠ B = 90 °, M is the midpoint of AC, D.E is the moving point on AB and BC respectively, and BD = CE, the shape of △ DEM is obtained and proved
- 10. In the triangle ABC, BD = DC = AC, D is the midpoint of BC and E is the midpoint of DC. It is proved that the bisection angle BAE of AD
- 11. In the triangle ABC, the angle ACB = 90 degrees, ad bisects the angle BAC, De is perpendicular to e, EF intersects AC to F, and EC intersects ad to o
- 12. In △ ABC, ∠ ACB = 100 °, AC = BC, point D is on AB, and BD = BC, point E is on AC, and AE = ad, EF ⊥ DC is on f 1. Find the degree of ∠ def 2: If AC = BC is removed from the title, what about the degree of ∠ def 3: If the body weight ∠ ACB = 100 ° is changed to ∠ ACB ＞∠ a, other things remain unchanged, what is the relationship between ∠ def and ∠ ACB
- 13. It is known that, as shown in the figure, in right angle trapezoid ABCD, ad is parallel to BC, ∠ ABC = 90 °, de ⊥ AC is at point F and BC is intersected at point G, which is called ab Verification: BG = FG No picture. Sorry
- 14. As shown in the figure, the vertex angle of the isosceles △ ABC is 50 ° and ab = AC. taking AB as the diameter, make a semicircle and intersect BC at point D and AC at point E BD、 De and The degree of the center angle to which AE is directed
- 15. Let d be a point on the edge BC of △ ABC, and AB2 = ad2 + BD × DC
- 16. As shown in the figure, △ ABC is an equilateral triangle, and points D, e and F are points on the segments AB, BC and Ca respectively, (1) If ad = be = CF, is △ def an equilateral triangle? Try to prove your conclusion; (2) If △ DEF is an equilateral triangle, is ad = be = CF true? Try to prove your conclusion
- 17. As shown in the figure, in the triangle ABC, ∠ 1 = ∠ 2, G is the midpoint of AD, lengthening BG intersecting AC with E.F is a point on AB, CF ⊥ ad is in H (1) Ad is the angular bisector of the triangle Abe; (2) Be is the center line on the side ad of the triangle abd; (3) Ch is the height on the edge ad of the triangle ACD A. 1 B. 2 C. Three D. 0
- 18. It is known that, as shown in the figure, △ ABC is an equilateral triangle, AE = CD, BQ ⊥ ad at Q, be crossing ad at point P, Confirmation: BP = 2pq
- 19. It is known that, as shown in the figure, in △ ABC, ad and AE are the bisectors of the height and angle of △ ABC respectively, if ∠ B = 30 ° and ∠ C = 50 ° (1) Find the degree of ∠ DAE; (2) What is the relationship between ∠ DAE and ∠ C - ∠ B? (no need to prove)
- 20. It is known that AD and AE are the angular bisector and height of triangle ABC respectively, ∠ B = 50 °, ead = 10 ° and ∠ C=____