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In the parallelogram ABCD, what is the length of CE if the extension line of AE bisector angle bad and intersecting DC is at point E, AB equals 4 and ad equals 9?
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The bisector of parallelogram ABCD, ∠ bad intersects straight line BC at e, intersects straight line DC at F, ∠ ABC = 120 ° FG / / CE and FG = C
The bisector of parallelogram ABCD, ∠ bad intersects the straight line BC at e, intersects the straight line DC at F, ∠ ABC = 120 ° FG / / CE and FG = CE, find the degree of ∠ BDG (need to draw your own drawing)
Ad = DF, EC = CF, ab = be connect eg and BG, because FG / / CE and FG = CE, GECF is diamond
In ▱ ABCD, the bisector BC of ∠ bad intersects at point E, and the bisector DC intersects at point F. if ∠ ABC = 120 ° FG ‖ CE, FG = CE, connect dB, DG, BG respectively, the size of ∠ BDG is ()
A. 30°B. 45°C. 60°D. 75°
Extend the intersection of AB and FG at h, connect HD. ∵ ad ∥ GF, ab ∥ DF, ∵ quadrilateral ahfd is parallelogram, ∵ - ABC = 120 °, AF bisects ∵ bad, ∵ DAF = 30 °, ∵ ADC = 120 °, ∵ DFA = 30 °, ∵ DAF is isosceles triangle, ∵ ad = DF, ∵ parallelogram ahfd is diamond, ∵ ADH
In the parallelogram ABCD, the bisector of angle bad intersects the straight line BC at point E and the straight line DC at point F. if ∠ ABC = 90 °, G is the midpoint of EF. Calculate the degree of ∠ BDG
∵ ABCD is a parallelogram, ABC = 90 °, ABCD is a rectangle,
∴∠BAD＝∠ECF＝90°、AB＝DC.
∵ - bad = 90 °, ∵ - BAE = ∵ bad / 2, ∵ - BAE = 45 ° and ∵ Abe = 90 ° and ∵ AB = be
From ab = DC, ab = be, DC = be
∵∠ BAE = 45 °, ab = be, ∵ AEB = 45 °, ∵ CEF = 45 ° and ∵ ECF = 90 ° and ∵ CE = CF
From DC = be, CF = CE, DC + CF = be + CE, DF = BC
∵∠ECF＝90°、EG＝FG,∴CG＝FG、∠DFG＝∠BCG＝45°.
From DF = be, FG = CG, ∠ DFG = ∠ BCG, we can get: △ DFG ≌ △ BCG, ∠ CDG = ∠ CBG,
The results show that B, G, C and D are in the same circle, BDG = BCG = 45 degree
RELATED INFORMATIONS
- 1. As shown in the figure, in the parallelogram ABCD, e and F are two points on the diagonal BD, and be = DF, connecting AE, AF, CE and cf. what kind of quadrilateral is the quadrilateral aecf, which shows your truth
- 2. As shown in the figure, in the parallelogram ABCD, AE bisects ∠ bad, CF bisects ∠ BCD, points E and F are on BD, connecting AF and CE, indicating that the quadrilateral aecf is parallel How else can you make a diagonal BD with two points e, f being quadrilateral and aecf being parallelogram
- 3. In known quadrilateral ABCD, angle B = angle d = 90 degree, angle AEC = angle bad, explain the reason of AE flat DC KKKK
- 4. As shown in the figure, in ▱ ABCD, it is known that m and N are the midpoint of AB and DC respectively. It is proved that a quadrilateral bmdn is also a parallelogram
- 5. As shown in the figure, in the quadrilateral ABCD, points E and F are two points on the diagonal, and be = DF. ① if the quadrilateral aecf is a parallelogram, is the quadrilateral ABCD a parallelogram? Why? ② if the quadrilateral aecf is a diamond, is the quadrilateral ABCD a diamond? Why?
- 6. As shown in the figure, it is known that in ▱ ABCD, e and F are two points on the diagonal BD, be = DF, points g and H are on the extension line of Ba and DC respectively, and Ag = ch connects Ge, eh, HF and FG
- 7. Given that a + B + C = 0, the absolute value of a = 3, the absolute value of B = 5, and the absolute value of C = 7, find the cosine value of the cosine of the cosine angle between the vector 2A + 3b and the vector 3a-b
- 8. Why is the absolute value of a plus B equal to the absolute value of a minus B when a is perpendicular to B,
- 9. Sinacosa = 60 of minus 169 Sina + cosa = 7 of minus 13 A is the second quadrant angle. How can we get Sina = 5 of minus 13 cosa = 1 of minus 13
- 10. As shown in the figure, in the triangular pyramid p-abc, PA = Pb = PC, if PA ⊥ Pb, PA ⊥ PC, Pb ⊥ PC, find the angle between PA and plane ABC
- 11. In the parallelogram ABCD, is the midpoint of e-type CD, a point on the edge of F-type BC, and the angle FAE = angle ead, EF and AE perpendicular? Give reasons
- 12. As shown in the figure, in the parallelogram ABCD, e is the midpoint of AD, the intersection of the extension line of CE and Ba and the point F, if BD = 2CD, the verification: ∠ f = ∠ BCF
- 13. In the parallelogram ABCD, ab = KBC, point P is on the diagonal BD, and angle ∠ ape = ∠ bad, PE and BC are compared with point E, and the quantitative relationship between PA and PE is explored
- 14. The edge length of rectangular paper ABCD is ab = 4, ad = 2. Fold the rectangular paper along EF so that point a and point C coincide. After folding, color it on one side (as shown in the figure), the area of the colored part is () A. 8B. 112C. 4D. 52
- 15. (Mianyang, 2011) as shown in the figure, fold a rectangular piece of paper ABCD with a length of 8cm and a width of 4cm so that point a and point C coincide, and the length of crease EF is equal to 2525cm
- 16. When a rectangular piece of paper ABCD is folded along EF, the intersection points of ED and BC are g, D and C at M and N, respectively. If ∠ EFG = 55 °, then ∠ 1=______ °，∠2=______ °．
- 17. As shown in the figure, after a rectangular piece of paper ABCD is folded along EF, points D and C respectively fall at d 'C', the intersection of ED 'and BC is g, if ∠ EFG = 65 degrees Find the degree of ∠ 1 and ∠ 2
- 18. As shown in the figure, fold a square piece of paper ABCD along EF, AD / / BC. If the angle EFG = 52 °, calculate the degree of angle deg
- 19. As shown in the figure, in the parallelogram ABCD, points E and F are on CD and ab respectively, and DF / / BF is on the point
- 20. As shown in the figure, it is known that in the parallelogram ABCD, e and F are respectively the midpoint of AD, G and H are two points on the diagonal BD, and BG = DH, what conclusion can be obtained?