Known: △ ABC ≌ △ EDC. Verification: be = ad

RELATED INFORMATIONS

• 1. As shown in the figure, the known points E and C are on the line BF, be = CF, ab ‖ De, ∠ ACB = ∠ F
• 2. Given ABC congruent triangle def, ab = 5, s triangle ABC = 10, find the height of de side in triangle def
• 3. If △ ABC ≌ Δ DEF is known, and the maximum angle of △ ABC is 100 °, then the maximum angle of △ DEF is______ ．
• 4. Given △ ABC ≌ △ def, and ∠ a = 52 °, B = 31 °, de = 10cm, if ∠ f = ∠ C, find the degree of ∠ F and the length of ab
• 5. Given △ ABC ≌ △ def, and ∠ a = 52 & # 186;, ∠ B = 31 & # 186;, ed = 10cm, if ∠ f = ∠ C, find the degree of ∠ F and the length of ab
• 6. As shown in the figure, triangle ABC is equal to triangle def, and angle a is equal to 52%, angle B = 31.21'ed = 1ocm
• 7. It is known that ab ‖ De, BC ‖ EF, D, C are on AF, and ad = CF
• 8. Given; point b.e.c.f is on the agreed line, ab = De, angle a = angle D, AC bisects DF; triangle ABC congruent triangle def, be = CF
• 9. As shown in the figure, points B, D, C, f are on a straight line, ab = De, angle a = angle D, AC is parallel to DF, prove 1, triangle ABC is equal to def 2, be = CF
• 10. The double length method of the central line of congruent triangle: ad is the central line of triangle ABC, e and F are on AB and AC respectively, and De is perpendicular to DF, then what is the relationship among be, CF and ef?
• 11. As shown in the figure, given that AD and be are the heights of △ ABC, ad and be intersect at point F, and ad = BD, can you find the congruent triangle in the figure? If you can find it, please explain why
• 12. As shown in the figure, the triangles ABC, congruent triangles def, am and DN are respectively the bisectors of the angles of the triangles ABC and def
• 13. Given three points a (0,4) B (- 3,0) C (3,0), now draw a parallelogram with ABC as the vertex. Please draw a figure according to the coordinates of three points and write out the coordinates of D
• 14. If EF = de + BF, then ∠ ECF = (). How do you get it, please give a brief introduction There is a little E on the edge ad of square ABCD, and a little f on ab. if EF = de + BF, then ∠ ECF = (). How do you get it, please give a brief explanation Don't be fussy when explaining, try to use mathematical laws and symbols (I'm junior one, please use the rotation change of junior one to solve this problem)
• 15. E. If f is the trisection of the hypotenuse of ABC, then Tan angle EFC =? I'm sorry. I'm asking for tanecf
• 16. If P and Q are the triangles of the hypotenuse BC, then Tan ∠ PAQ =?
• 17. Δ ABC is isosceles right triangle, ∠ BAC = 90 °, D is a point on AC, extend Ba to e, make AE = ad, prove BD perpendicular to CE
• 18. In △ ABC, ab = AC, CG ⊥ Ba intersects the extension line of BA at point g. an isosceles right triangle ruler is placed as shown in Figure 1. The right angle vertex of the triangle ruler is F. one right angle side is in a straight line with AC side, and the other right angle side just passes through point B (1) In Figure 1, please guess and write down the quantitative relationship between BF and CG by observing and measuring the length of BF and CG, and then prove your conjecture; (2) when the triangle ruler moves to the position shown in Figure 2 along the direction of AC, one right angle side is still on the same line with AC side, the other right angle side intersects BC side at point D, and makes de ⊥ BA at point e through point D. at this time, please observe and measure De, D The length of F and CG, conjecture and write the quantitative relationship between de + DF and CG, and then prove your conjecture; (3) whether the conjecture in (2) is still true when the triangle ruler continues to move to the position shown in Fig. 3 along the direction of AC on the basis of (2) (the point F is on the line AC, and the point F does not coincide with point C) (do not explain the reason)
• 19. The △ ACB is an isosceles triangle, ∠ ACB = 90 ° extend Ba to e, extend AB to F, ∠ ECF = 135 ° Q: is there an equivalent relationship between AB, AE and BF
• 20. As shown in the figure, △ ABC is an isosceles triangle, ∠ ACB = 90 °, extend AB to F, so that ∠ ECF = 135 °. Verification: AE: EC = Ba: CF