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As shown in the figure, given that point E is on the edge ab of △ ABC, point D is on the extension line of Ca, and point F is on the extension line of BC, what is the size relationship between ∠ ACF and ∠ D? Please give reasons
The reason is as follows: ∵ BAC is the outer angle of △ ade, ∵ BAC ＞ D, ∵ ACF is the outer angle of △ ABC, ∵ ACF ＞ BAC, so ∵ ACF ＞ D
As shown in the figure, given that point E is on the edge ab of △ ABC, point D is on the extension line of Ca, and point F is on the extension line of BC, what is the size relationship between ∠ ACF and ∠ D? Please give reasons
The reason is as follows: ∵ BAC is the outer angle of △ ade, ∵ BAC ＞ D, ∵ ACF is the outer angle of △ ABC, ∵ ACF ＞ BAC, so ∵ ACF ＞ D
As shown in the figure, given that point E is on the edge ab of △ ABC, point D is on the extension line of Ca, and point F is on the extension line of BC, what is the size relationship between ∠ ACF and ∠ D? Please give reasons
The reason is as follows: ∵ BAC is the outer angle of △ ade, ∵ BAC ＞ D, ∵ ACF is the outer angle of △ ABC, ∵ ACF ＞ BAC, so ∵ ACF ＞ D
As shown in the figure, BD and CE are the height of edge AC and edge ab of triangle ABC respectively, BD = CE, is segment EB equal to segment CD? Why?
EB is equal to the segment CD
In RT △ BDC and RT △ CEB,
ce=bd,bc=bc,
Then RT △ BDC ≌ RT △ CEB,
cd=be.
As shown in the figure, B = C, BD = CE, CD = BF
Another question
Because AB = CD, ad = BC, ABCD is a parallelogram
So AF is parallel to CE, so angle f equals angle E
The points c, D, e are on the line ab. given that ab = 12cm, CE = 4cm, find the length sum of all the line segments in the graph
Line segment order a_____ C___ D___ E_____________ B
Length sum of all segments = AC + AD + AE + AB + CD + CE + CB + de + DB + EB
=(AC+CD+DE+EB)+(AD+DB)+AB+AE+CE+CB
=AB+AB+AB+(AC+CE)+CE+CB
=3AB+(AC+CB)+2CE
=4AB+2CE
=48+8=56
Extend line AB to point C so that BC = one third of AB, extend AC to point D in reverse so that ad = half of AC, then CD = --- ab
CD=2AB.
Because BC = 1 / 3AB, so AC = 4 / 3AB, so ad = 2 / 3AB;
So CD = AD + AB + BC = 2 / 3AB + AB + 1 / 3AB = 2Ab
Extend line AB to C to make BC = 12ab, extend AC to D reversely to make ad = 12ac, if AB = 8cm, then CD=______ cm．
As shown in the figure, BC = 12ab = 4, AC = AB + BC = 8 + 4 = 12cm, ad = 12ac = 6, CD = AD + AC = 12 + 6 = 18cm
CD is two points on the line AB, known AC: CD: DB = 2:3:4, points E and F are the midpoint of AC and BD respectively, and EF = 5cm, find the length of ab
Let AC = 2x, then CD = 3x, DB = 4x, ab = AC + CD + DB = 9x,
Because points E and F are the midpoint of AC and BD respectively, then EC = AC / 2 = x, DF = dB / 2 = 2x,
So EF = EC + CD + DF = 6x = 5, that is, x = 5 / 6, so AB = 9x = 15 / 2 (CM)
As shown in the figure, C and D are two points a and B on the line segment, known as AC: CD: DB = 2:3:4, point EF is the midpoint of AC and BD respectively, and EF = 5cm, find the length of ab
Let EA = x, ab = 2Y, FD = Z
x/z=2:4
x/2y=2:3
ef=
>>>>
z=2x
x=4/3y
x+2y+z=5
>>
x=4/3y
3x+2y=5
>>
y=1.2
>>
x=1.6
>>
z=3.2
>>
ab=(x+y+z)*2=12
RELATED INFORMATIONS
- 1. As shown in the figure, given that point E is on the edge ab of △ ABC, point D is on the extension line of Ca, and point F is on the extension line of BC, what is the size relationship between ∠ ACF and ∠ D? Please give reasons
- 2. As shown in the figure, the triangle ABC is an equilateral triangle, and D is the midpoint on the side ab. it is known that the area of the triangle BDE is 5 square centimeters. Find the area of the equilateral triangle ABC
- 3. In △ ABC, ab = AC, be = CF, EF intersect BC with D
- 4. As shown in the figure, it is known that ad, be and CF are respectively the heights of the three sides of △ ABC, h is the perpendicular, and the extension line of ad intersects the circumscribed circle of △ ABC at point G
- 5. As shown in the figure, in the quadrilateral ABCD, ∠ bad = ∠ ABC, ad = BC, AC intersects with point o at BD. let's say that the triangle OAB is an isosceles triangle
- 6. As shown in Figure 1, in △ ABC, points e.d.f ab.bc.ca If ad bisects ∠ BAC, then the quadrilateral AEDF is a diamond?
- 7. In the triangle ABC, ab = BC, the point D E F is the midpoint on the side of BC AC AB respectively
- 8. In the geometry shown in the figure, EA vertical plane ABC, DB vertical plane ABC, AC vertical plane BC and AC = BC = BD = 2ae. M is the midpoint of ab Finding the tangent of the angle between de and plane EMC
- 9. As shown in the figure, in △ ABC, points D and E are on AB and AC respectively, de ‖ BC, ∠ CED = ∠ BDC. (1) prove: △ DCE ∽ CBD; (2) if BC = 2CD, s △ ade = 1, find the value of s △ ABC
- 10. Help out on the inverse proportion function and Pythagorean theorem of the problem it! A little basic on the line, but not too simple!
- 11. In △ ABC, ∠ C = 90 °, AC = 4cm, BC = 5cm, point D is on BC, and CD = 3cm. There are two moving points P and Q starting from point a and point B at the same time, where point P moves along AC to terminal C at the speed of 1cm / s, and point Q moves along BC to terminal C at the speed of 1.25cm/s. Passing point P, PE ‖ BC is connected to point E, and EQ is connected (1) The length of AE and De is expressed by the algebraic expression containing x; (2) when the point Q moves on BD (excluding points B and D), let the area of △ EDQ be y (cm2), find the functional relationship between Y and time x, and write out the value range of the independent variable x; (3) when the value of X is, the △ EDQ is a right angle triangle
- 12. In RT △ ACB, ∠ ACB = 90 °, AC = 3, BC = 4, the moving circle O passing through point C is tangent to the hypotenuse AB at the moving point P, and the value range and maximum value of CP are obtained
- 13. In a convex quadrilateral ABCD, Da = DB = DC = BC, then the degree of the largest angle in the quadrilateral is () A. 120°B. 135°C. 150°D. 165°
- 14. As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 108 °, D is on AC and BC = AB + CD, the proof is: BD bisects ∠ ABC
- 15. As shown in the figure, in the isosceles trapezoid ABCD, it is known that ad ∥ BC, ab = CD, AE ⊥ BC are in E, ∠ B = 60 °, DAC = 45 ° and AC = 6?
- 16. As shown in the figure, in RT △ ABC, ∠ C = 90 °, am is the middle line on the side of BC, sin ∠ cam = 35, then the value of tanb is () A. 32B. 23C. 56D. 43
- 17. In rectangular ABCD, AE is perpendicular to BD, and the angle DAE = 2 angle BAE?
- 18. Ad is the height of triangle ABC, be is the midline angle on the side of AC, CBE = 30 degrees to prove ad = be
- 19. As shown in Figure 7-11, in the known triangle ABC, ad is perpendicular to D, BC is equal to D, AE is equal to ∠ BAC (2), if ∠ B ＞ C, try to explain ∠ DAE = 1 / 2 (∠ B - ∠ C)
- 20. It is known that in the oblique triangle ABC, the straight line with high BD and CE intersects at h, and the angle a = 45 degrees, so the degree of the angle BHC can be calculated? No picture, if you can,