Known: as shown in the figure, in trapezoidal ABCD, AB is parallel to CD, AC is vertical to BC, ad is vertical to BD, e is the midpoint of AB, prove: angle ECD is equal to angle EDC
It's very simple. You can see that CE and de are the middle lines on the hypotenuse of two right triangles,
The center line on the hypotenuse of a triangle is equal to half of the hypotenuse, that is, CD = CE = 1 / 2Ab
So the two angles are equal
RELATED INFORMATIONS
- 1. In the quadrilateral ABCD, BD is perpendicular to ad, AC is perpendicular to BC, e is the key point of AB, and the verification angle EDC = angle ECD Wait online. It's urgent
- 2. As shown in the figure, in the quadrilateral ABCD, it is known that ∠ a = ∠ B = 90 ° e picks up the midpoint of AB, ∠ EDC = ∠ ECD, and proves that the quadrilateral ABCD is a rectangle. As shown in the figure (the figure can't be uploaded, but it is roughly a quadrilateral that looks like a rectangle, and the midpoint e connects the fixed points D and C to form a triangle)
- 3. It is known that in the quadrilateral ABCD, AD / / BC, angle BAC = angle D, and point E.F BC.CD And QE AEF = angle ACD
- 4. In trapezoidal ABCD, ad ‖ BC, ab = AC, if ∠ d = 110 °, ACD = 30 °, then ∠ BAC is equal to () A. 80°B. 90°C. 100°D. 110°
- 5. In the isosceles trapezoid ABCD, the extension lines of AD / / BC, de / / AC and BC intersect at the point E, CA bisects ∠ BCD, and proves ∠ B = 2 ∠ E
- 6. As shown in the figure, if the quadrilateral ABCD is trapezoid, ad ‖ BC, CA is bisector of ∠ BCD, and ab ⊥ AC, ab = 4, ad = 6, then tanb = () A. 23B. 22C. 114D. 554
- 7. As shown in the figure, in the right angle trapezoid ABCD, ad ‖ BC, ab ⊥ BC, ∠ DCB = 75 °, the other vertex e of equilateral △ DCE with CD as one side is on the waist ab. (1) calculate the degree of ∠ AED; (2) prove: ab = BC
- 8. As shown in Figure 1, in the rectangular trapezoid ABCD, ad ‖ BC, ab ⊥ BC, ∠ DCB = 75 ° and the other vertex e of equilateral △ DCE with CD as one side is on the waist ab (1) (2) AB = BC; (3) as shown in Figure 2, if f is a point on the line CD, ∠ FBC = 30 °, the value of dffc can be obtained
- 9. As shown in the figure, in the right angle trapezoid ABCD, ad ‖ BC, ab ⊥ BC, ∠ DCB = 75 °, the other vertex e of equilateral △ DCE with CD as one side is on the waist ab. (1) calculate the degree of ∠ AED; (2) prove: ab = BC
- 10. In trapezoidal ABCD, AD / / BC, ab = CD = 2, BC = 6, point E is on BD, and angle DCE = angle ADB 1) Find out all the similar triangles in the graph and prove them; 2) Let BD = x, be = y, find the analytic expression of Y with respect to x, and write out its domain of definition; 3) When ad = 4, find the length of be
- 11. In rectangular ABCD, ab = 4, ad = 2, point m is the midpoint of AD. point E is a moving point on edge ab. connect EM and extend the intersection line CD at point F, cross m to make the vertical line of EF, the extension line of BC at point G, connect eg, and the intersection side DC at point Q. let the length of AE be x and the area of triangle EMG be y (1) Find the tangent value of ∠ MEG; (2) Find the analytic expression of Y with respect to x, and write out the value range of X; (3) The midpoint of Mg is denoted as point P and connected with CP. if {PGC} efq, find the value of Y
- 12. As shown in the figure, in rectangle ABCD, ab = 3, BC = 4, e is the moving point on edge ad, f is a point on ray BC, EF = BF and intersecting ray DC at point G, let AE = x, BF = y (1) When △ bef is an equilateral triangle, find the length of BF (2) Find the analytic expression of the function between Y and X, and write out its domain of definition (3) Fold △ Abe along the straight line be, and point a falls at a '. Try to explore whether △ a'bf can be an isosceles triangle? If so, ask for the length of AE; if not, explain the reason As long as the answer and steps of the third question
- 13. Trapezoid ABCD is right angle trapezoid, AB / / DC, ∠ DAB = 90 °, PA ⊥ plane ABCD, and PA = ad = DC = & frac12; ab = 1, M is the midpoint of Pb, find the dihedral angle formed by plane AMC and plane BMC
- 14. As shown in the figure, in the trapezoidal ABCD, AD / / BC, ∠ ABC = 60 °. ∠ DCB = 30 ° AB = 4, find BC ad
- 15. In rectangular trapezoid ABCD, ab ∥ CD, ad ⊥ DC, ab = BC, and AE ⊥ BC
- 16. As shown in the figure, in rectangular trapezoid ABCD, ab ∥ CD, ad ⊥ DC, ab = BC, and AE ⊥ BC. (1) prove: ad = AE; (2) if ad = 8, DC = 4, find the length of ab
- 17. As shown in the figure, in rectangular trapezoid ABCD, ab ∥ CD, ad ⊥ DC, ab = BC, and AE ⊥ BC. (1) prove: ad = AE; (2) if ad = 8, DC = 4, find the length of ab
- 18. It is known that in the trapezoidal ABCD, ad ‖ BC, AB: BC: CD: Da = 4:5:3:2, if BC-AD = 9, the lengths of AB and DC are obtained
- 19. As shown in the figure, in the rectangular trapezoid ABCD, the bottom AB = 13, CD = 8, ad ⊥ AB and ad = 12, then the distance from a to BC is () A. 12B. 13C. 12×2113D. 10.5
- 20. In the right angle trapezoid ABCD, ab ∥ CD, Da is vertical AB, ab = 13, CD = 8, ad = 12, then what is the distance from point a to BC?