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As shown in the figure, it is known that: in trapezoidal ABCD, ad ‖ BC, e is the midpoint of AC, connecting de and extending BC to point F, connecting AF. (1) verification: ad = CF; (2) under the condition that the original condition remains unchanged, please add another condition (no additional auxiliary line), so that the quadrilateral AFCD becomes a diamond, and explain the reason
(1) It is proved that in △ DEA and △ FEC, ∫ ad ≌ BC, ∫ DAE = ∠ FCE, ∫ ade = ∠ efc. And ∫ e is the midpoint of AC, ∫ AE = CE. ≌ DEA ≌ FEC. ∫ ad = cf. (2) Da = DC is added. It is proved that ∫ ad ∥ BC, ∫ ad = CF, ∫ quadrilateral AFCD is parallelogram, and ∫ Da = DC, ≌ quadrilateral AFCD is diamond
RELATED INFORMATIONS
- 1. In the trapezoid ABCD, ad ∥ BC, BF = FG = GC, it is proved that AP: FP = AF: ef (P is a point under the bottom edge of the trapezoid, AP intersects at point F through BC, DP intersects at point G through BC, and the intersection of BD and AP is e) BC is a trapezoid. The bottom (long) edge of ABCD is below the trapezoid, that is, BC
- 2. In trapezoidal ABCD, ab ‖ CD, f is the midpoint of BC, and AF ⊥ ad, e is on CD, satisfying AF = EF. (1) verification: 12 ∠ AFE + ∠ d = 90 °; (2) connecting AE, if ad = 5, AF = 6, finding the length of AE
- 3. As shown in the figure, in trapezoidal ABCD, ad ∥ BC, ab = DC, AE ⊥ BC at point E, the vertical bisector GF of AB intersects BC at point F, intersects AB at point G, and connects AF. it is known that ad = 1.4, AF = 5, GF = 4. (1) calculate the waist length of trapezoidal ABCD; (2) calculate the area of trapezoidal AFCD
- 4. In the isosceles trapezoid ABCD, AD / / BC, AB / / DC, point E is a point in the trapezoid, and EA = ed, EB = EC is proved
- 5. In the trapezoidal ABCD, ab ‖ CD, ∠ a = 90 °, ab = 2, BC = 3, CD = 1, e is the midpoint of AD. try to judge the position relationship between EC and EB, and write the inference
- 6. As shown in the figure, ladder ABCD, e is the midpoint of AB, DC = AD + BC, prove de ⊥ EC
- 7. In ladder ABCD, AD / / BC, DC / / BC, e are the midpoint of ab It's BC ⊥ DC
- 8. In trapezoidal ABCD, if ab ‖ CD, point E is the midpoint of AD and s △ BEC = 2, then the area of trapezoidal ABCD is______ ．
- 9. As shown in the figure, in trapezoidal ABCD, ∠ d = 90 ° and M is the midpoint of ab. if cm = 6.5 and BC + CD + Da = 17, the area of trapezoidal ABCD is () A. 20B. 30C. 40D. 50
- 10. As shown in the figure, in ladder ABCD, AB is parallel to DC, point E is the midpoint of DC, ∠ AED = ∠ BEC, proving that ladder ABCD is isosceles ladder
- 11. As shown in the figure, in ladder ABCD, AB is parallel to CD, ad = BC, EF are AD.BC BD bisection ∩ ABC is called EF at point G, eg = 18, FG = 10 to find the trapezoid perimeter
- 12. In trapezoid ABCD, the midpoint ad ‖ BC, ab = ad = DC = 5, cos ∠ ABC = 3 / 5, e is the midpoint of edge AB, and point F is the first moving point of ray BC, connecting BD and DF (1) Finding Tan angle abd when DF is perpendicular to BC (2) When point F is on the extension line of BC, connect EF with DC at point G, let CF be equal to m, and find the length of line segment DG (expressed by the algebraic expression containing m) (3) Let m be a point on DC and 5DM equal to 8ae. Join am to intersect diagonal BD at point n. if triangle BDF equals triangle and, find the length of segment CF
- 13. As shown in the figure, in square ABCD, P is on the diagonal BD, e is on the extension line of CB, and PE = PC, passing through point P as PF ⊥ a in F, and the straight line pf intersects AB, Cd in G and h respectively. (1) prove: DH = Ag + be; (2) if be = 1, ab = 3, find the length of PE
- 14. In square ABCD, P is a point on diagonal AC, PE ⊥ AB is in E, PF ⊥ BC is in F. try to guess the quantity and position relationship between EF and PD, and give the proof We need to prove the process in detail
- 15. In the square ABCD, P is a point on the diagonal AC, PE ⊥ AB, PF ⊥ BC, and the perpendicular feet are E.F PD.EF Of
- 16. As shown in the figure: in square ABCD, AC = 10, P is any point on AB, PE ⊥ AC is in E, PF ⊥ BD is in F, then PE + PF=______ It can be summed up in one sentence: the sum of the distances from any point on the side of a square to two diagonals is equal to______ ．
- 17. As shown in the figure, in rectangle ABCD, ab = 3, BC = 4, point P is the first moving point on diagonal AC, PE ⊥ pf intersects ad, AB with E and f respectively, proving: PE / PF = 3 / 4
- 18. In square ABCD, P is the moving point on ad, PE ⊥ AC is in E, PF ⊥ BD is in F, PE + PE = 5, then the circumference of square ABCD is?
- 19. It is known that in square ABCD, P is a point on diagonal AC, PE ⊥ AB, PF ⊥ BC, and the perpendicular feet are e and f respectively
- 20. In ABCD, ab = AC, CB = CD, point P is a point on diagonal AC, PE is perpendicular to BC and E, PF is perpendicular to CD and F, PE = PF is proved