![](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, we know ∠ 1 + ∠ 2 = 180 ° and ∠ 3 = ∠ B. try to judge the size relationship between ∠ AED and ∠ C, and rationalize the conclusion
It is proved that: ∵ 1 + ∠ 4 = 180 ° (definition of adjacent complementary angle)
∠ 1 + ∠ 2 = 180 ° (known)
Ψ 2 = ∠ 4 (the complementary angles of the same angle are equal)
/ / EF ∥ AB (the internal staggered angle is equal and the two lines are parallel)
Ψ 3 = ∠ ade (two lines are parallel, and the internal staggered angle is equal)
And ∵ B = ∵ 3 (known),
Ψ ade = ∠ B (equivalent substitution),
▽ de ‖ BC (equal position angle, two parallel lines)
Ψ AED = ∠ C (two lines are parallel and the same position angle is equal)
As shown in the figure, in the right triangle ABC. ∠ C = 90 ° ad bisects ∠ cab. And intersects with the bisector be of ∠ ABC at point D. find the degree of ∠ ade SOS
This is a calculation problem
∵RT△ABC.
∴∠CAB+∠ABC=90°.
And ∵ ad, be are the angular bisectors of ∵ cab and ∵ ABC respectively
∴1/2∠CAB+1/2∠ABC=45°.
That is, the sum of two internal angles of a triangle equals to the outer angle of the third triangle
As shown in the figure, AB parallels CD, AE bisects bad, CD and AE intersect at point F, ∠ CFE = ∠ e, verification: ad parallel BC This is the picture
∵ AB//CD
Ψ BAE = ∠ CFE (two lines are parallel, and the same position angle is equal)
AE bisection ∠ bad
∴∠BAE＝∠EAD
﹣ CFE = ∠ ead (equivalent substitution)
And ∵, ∵ CFE = ∠ e
Ψ ead = ∠ e (equivalent substitution)
/ / AD / / BC (the internal staggered angle is equal and the two lines are parallel)
Given that AB is parallel to CD, AE bisects angle bad, CD and AE intersect at point F, angle CFE = angle e, try to explain that ad is parallel to BC
∵ AE bisecting angle bad
∴∠BAE=∠DAE
∵AB‖CD
∴∠BAE=∠DFA
∵∠DFA=∠CFE=∠E
∴∠DAE=∠E
∴AD‖BC
Ad is parallel to BC, e is the midpoint of line CD, AE bisection angle bad, and be bisection angle ABC is proved
E is the midpoint of CD, ∵ f is the midpoint of AB, (EF is the median line of trapezoid)
Ad ∥ BC, ab ∥ BC, e is the midpoint of CD
In trapezoidal ABCD, a straight line parallel to BC is made through the midpoint e of CD and intersects AB with point F. according to the known conditions, f is the midpoint of ab. ∵ e is the midpoint of CD, EF / / BC / / ad ᙨ EF is the median line of trapezoid ABCD
As shown in the figure, ab ‖ CD, AE bisection ∠ bad, CD and AE intersect at F, ∠ CFE = ∠ E. verification: ad ‖ BC
Proof: ∵ AE bisection ∠ bad,
∴∠1=∠2，
∵AB∥CD，∠CFE=∠E，
∴∠1=∠CFE=∠E，
∴∠2=∠E，
∴AD∥BC．
In trapezoidal ABCD, ∠ d = 90 °, AB is parallel to CD, ab = BC = 20cm, DC = 4cm, AE is perpendicular to BC, then AE =? S trapezoid ABCD =? S?
Therefore, in the case of ⊥ B ⊥ B, the solution of ⊥ B ⊥ B = B ⊥ B ⊥ B = B ⊥ B, so that ⊥ B = B ⊥ B ⊥ B ⊥ B ⊥ B ⊥ B ⊥ B ⊥ B ⊥ so ⊥ B ⊥ B ⊥ B ⊥, so ⊥ B ⊥ B ⊥ B ⊥ B ⊥ B ⊥ B
In trapezoid ABCD, ∠ d = 90du, ab ‖ CD, ab = BC = 20cm, DC = 4cm, AE ⊥ BC, find AE =?, s trapezoid =? DC on top, AB on bottom, E on top
The correct answer is: AE = 12cms trapezoid = (4 + 20) * 12 / 2 = 144cm ^ 2 because this problem is to draw auxiliary lines, I have no tools to draw, not easy to say! First of all, cross point C to make a vertical line CF, vertical AB, intersect AB with point F. in this way, in the right triangle CFB, BC = 20cm, FB = 20-4 = 16cm. According to the Pythagorean theorem, CF = 12cms triangle ACB = AE * BC / 2 = CF * AB
In trapezoid ABCD, ab ‖ DC, ∠ d = 90 °, ab = BC = 20cm, DC = 4cm, AE ⊥ BC, calculate the length of AE and the area of ABCD
Make the vertical intersection of AB and F through point C
Since angle B = angle B angle e = angle o = 90 degree AB = BC, triangle Abe is all equal to triangle BOC
ob=ab-ao=ab-cd=20-4=16
AE = co = CB ^ 2-ob ^ 2 Radix = 8 Radix 6
Area = (CD + AB) * Co / 2 = 96 root number 6
RELATED INFORMATIONS
- 1. As shown in the figure, AB / / CD, ad vertical DC, ab = BC, and AE vertical BC 1. Prove that ad = AE If ad = 8, DC = 4, the length of AB is 4
- 2. (Mathematics) in RT △ ABC, ∠ B = 90 degrees, BC > AB, BP = ba, BP=BA,BD⊥AC,PE⊥AC In the four line segments ad, BD, De, PE, there is a certain quantitative relationship between some line segments. Please write an equation to express the quantitative relationship (the equation contains 2 or 3 line segments), and explain the reasons for the equation
- 3. In the triangle ABC, the angle c = 90 degrees, AC = BC, P is any point of AB, PE is vertical BC, PF is vertical AC, M is the midpoint of AB, connecting me, MF It is proved that me = MF
- 4. Given that the triangle ABC, P is a point in the triangle, through P as PD vertical BC, PE vertical AC, PF vertical AB, the height of triangle ABC h, it is proved that PD + PE + pf = H
- 5. In the RT triangle ABC, ∠ C = 90 ° C = 34. If a to B is equal to 8 to 15, then a = b =
- 6. In Baidu RT triangle ABC, angle c = 90 degrees, angle a = 30 degrees, AC = 2
- 7. Find the S-T, V-T images of the intermediate velocity and the intermediate displacement velocity
- 8. What do acceleration and time images show Speed and time indicate acceleration. Distance and time are speed. What about acceleration and time?
- 9. The relationship between physical problems (displacement of uniform linear motion) and velocity The speed of a car before braking is 90 km / h, and the acceleration obtained by braking is 10 m / S 2 (1) The sliding distance within 10s after the start of braking Thank you very much!
- 10. It is proved that the instantaneous velocity at the midpoint of displacement is VX / 2 = √ VO2 + vt2 / 2 in the linear motion with uniform speed change
- 11. As shown in the figure, in △ ABC, ∠ ABC = ∠ C = 2 ∠ a, BD ⊥ AC intersects AC at point D, then ∠ DBC=______ .
- 12. As shown in the figure, in △ ABC, D is a point on the edge of BC, ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, ∠ BAC = 63 ° to find the degree of angle DAC
- 13. The lines AB and CD intersect at o.oe and of are bisectors of angle AOC and angle BOD respectively. (1) draw the figure. (2) ray OE The lines AB and CD intersect at points o.oe and of are bisectors of angle AOC and angle BOD respectively. (1) draw this figure. (2) are ray OE and of on the same line? (3) draw bisector of angle AOD OG.OE What is the location relationship with og?
- 14. As shown in the figure, the straight lines a and B are cut by the lines C and D, ∠ 1 and ∠ 2 are complementary angles to each other, ∠ 3 = 117 ° and the degree of ∠ 4 is calculated
- 15. As shown in the figure, in trapezoidal ABCD, ad ‖ BC, ∠ a = 90 °, e is the point above AB, EC = ed, ∠ BEC = 75 °, AED = 45 °, verification: ab = BC
- 16. As shown in the figure, is the line AB parallel to CD? And the line AD and BC? Why? Because of what, so what
- 17. As shown in the figure, AC bisects ∠ DAB, ab ＞ ad, CB = CD, CE ⊥ AB in E, (1) Results: ab = AD + 2EB; (2) If ad = 9, ab = 21, BC = 10, find the length of AC
- 18. In the triangle ABC, if the angle c is equal to 90 degrees, AC is equal to 8 times the root sign 3, and ab is equal to 10 times the root sign 3, what is the value of Tana
- 19. In △ ABC, the opposite sides of angles a, B and C are a, B, C. 2Sin? C = 3cosc, C = √ 7, and the area of △ ABC is (3 √ 3) / 2, ① The size of angle c; ② The value of a + B
- 20. In the triangle ABC, the angle B = 90 ', the two right sides AB = 7, BC = 24. There is a point P in the triangle whose distance from each side is equal. What is the distance?