![](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>)
It is known that in △ ABC, ab = AC, BD and CE are the heights on the edge of AC and ab respectively, connecting De Results: (1) △ abd ≌ △ ace; (2) The quadrilateral BCDE is isosceles trapezoid
It is proved that: (1) BD and CE are the heights of AC and ab respectively, and ∵ a = a, ab = AC, abd ≌ △ ace; (2) if ad = AE is obtained from △ abd ≌ △ ace, then ∵ ade = 180 ° − A2. ∵ AB = AC, so ? ACB = 180 ° − A2
It is known that: as shown in the figure, in the isosceles triangle ABC, ab = AC, BD ⊥ AC, CE ⊥ AB, and the perpendicular feet are points D and E, respectively, to verify; be = CD
prove:
∵BD⊥AC,CE⊥AB
∴∠ADB＝∠AEC＝90
∵AB＝AC,∠BAD＝∠CAE
∴△ABD≌△ACE （AAS）
∴AD＝AE
∵BE＝AB-AE,CD＝AC-AD
∴BE＝CD
As shown in the figure, in △ ABC, BD and CE are the heights on the sides of AC and ab respectively. If BD = CE, then △ ABC is an isosceles triangle. Why?
It is proved that ∵ BD and CE are higher than ∵ ABC,
△ BCD and △ CBE are right triangles,
In RT △ BCD and RT △ CBE,
BC＝CB
BD＝CE ，
∴Rt△BCD≌Rt△CBE（HL），
∴∠ABC=∠ACB，
∴AB=AC，
That is, △ ABC is an isosceles triangle
As shown in the figure, in △ ABC, ∠ B = 90 °, ab = BC, BD = CE, M is the midpoint of AC edge, Verification: △ DEM is isosceles triangle
Proof: connect BM,
Because AB = BC, am = MC,
So BM ⊥ AC, and ∠ ABM = ∠ CBM = 1
2∠ABC=45°，
Because AB = BC,
So ∠ a = ∠ C = 180 ° − ABC
2=45°，
So ∠ a = ∠ ABM, so am = BM,
Because BD = CE, ab = BC, so ab-bd = bc-ce, that is, ad = be,
In △ ADM and △ BEM,
AD＝BE
∠A＝∠EBM＝45°
AM＝BM ，
So △ ADM ≌ △ BEM (SAS),
So DM = em,
So △ DEM is an isosceles triangle
As shown in the figure, in △ ABC, ∠ B = 90 °, ab = BC, BD = CE, M is the midpoint of AC edge, Verification: △ DEM is isosceles triangle
It is proved that: connecting BM, because AB = BC, am = MC, so BM ⊥ AC, and ∠ ABM = ∠ CBM = 12 ∠ ABC = 45 ° because AB = BC, so ∠ a = ∠ C = 180 ° − ABC2 = 45 °, so ∠ a = ∠ ABM, so am = BM, because BD = CE, ab = BC, so ab-bd = bc-ce, that is ad = be, in △ ADM and △ BEM
As shown in the figure, in △ ABC, ∠ B = 90 °, ab = BC, BD = CE, M is the midpoint of AC edge, Verification: △ DEM is isosceles triangle
It is proved that: connecting BM, because AB = BC, am = MC, so BM ⊥ AC, and ∠ ABM = ∠ CBM = 12 ∠ ABC = 45 ° because AB = BC, so ∠ a = ∠ C = 180 ° − ABC2 = 45 °, so ∠ a = ∠ ABM, so am = BM, because BD = CE, ab = BC, so ab-bd = bc-ce, that is ad = be, in △ ADM and △ BEM
As shown in the figure, in △ ABC, ∠ B = 90 °, ab = BC, BD = CE, M is the midpoint of AC edge, Verification: △ DEM is isosceles triangle
It is proved that: connecting BM, because AB = BC, am = MC, so BM ⊥ AC, and ∠ ABM = ∠ CBM = 12 ∠ ABC = 45 ° because AB = BC, so ∠ a = ∠ C = 180 ° − ABC2 = 45 °, so ∠ a = ∠ ABM, so am = BM, because BD = CE, ab = BC, so ab-bd = bc-ce, that is ad = be, in △ ADM and △ BEM
As shown in the figure, in △ ABC, ∠ B = 90 °, ab = BC, BD = CE, M is the midpoint of AC edge, Verification: △ DEM is isosceles triangle
Proof: connect BM,
Because AB = BC, am = MC,
So BM ⊥ AC, and ∠ ABM = ∠ CBM = 1
2∠ABC=45°，
Because AB = BC,
So ∠ a = ∠ C = 180 ° − ABC
2=45°，
So ∠ a = ∠ ABM, so am = BM,
Because BD = CE, ab = BC, so ab-bd = bc-ce, that is, ad = be,
In △ ADM and △ BEM,
AD＝BE
∠A＝∠EBM＝45°
AM＝BM ，
So △ ADM ≌ △ BEM (SAS),
So DM = em,
So △ DEM is an isosceles triangle
In the triangle ABC, ad is the center line, point E is on ad, AE = ed, connect CE and extend the intersection AB to point F, and find the quantitative relationship between AF and BF And explain the reasons
1:2
Make DP ∥ FC by D and pass BF to P
Because e is the midpoint of AD, AF = FP
And because D is the midpoint of BC, FP = Pb
So f is the trisection of ab
It is known that in △ ABC, ad is the midline, point E is on ad, AE = ed, CE is connected and Ba is extended to point F, what is the quantitative relationship between AF and BF? There is no picture
It is suggested that the midpoint P of CF is taken to connect DF, and it is proved that △ AFE and △ EDP are congruent, which can be obtained from the median line of the triangle. AF = 1 BF in 2
RELATED INFORMATIONS
- 1. It is known that ad is the median line of the triangle ABC, e is the extension of CE at any point on ad, and the intersection of AB and F is proved to be AE / ad = 2AF / BF
- 2. As shown in the figure, the triangle ABC and the triangle ECD are isosceles right triangles, in which AC = BC, DC = EC, and C on ad, connect AE, de. please find out a pair of congruent triangles in the graph, and write the process of proving their congruence
- 3. As shown in the figure, in RT triangle ABC, angle c = 90 °, angle 1 = equal to angle 2, CD = 1.5, BD = 2.5?
- 4. In the RT triangle ABC, if the angle ACB = 90 degrees, BC = 3, AC = 4, the vertical bisector De of AB intersects the extension of BC at point E, then
- 5. As shown in the figure, in △ ABC, ∠ ACB = 90 °, points D and E are on AB, ad = AC, be = BC. Try to judge whether the size of ∠ DCE is related to the degree of ∠ B. if so, ask for the relationship between them; if not, please determine the degree and explain the reason
- 6. As shown in the figure, in △ ABC, ab = AC, ad ⊥ BC at D, ﹤ bad = 40 °, point E on AC, and AE = ad, calculate the degree of ∠ CDE ditto
- 7. As shown in the figure, ab = AC, ad = AE are known
- 8. It is known that, as shown in the figure, △ ABC is an inscribed regular triangle of ⊙ o, and point P is a moving point on arc BC. This paper explores the relationship between PA, Pb and PC
- 9. The triangle ABC is an equilateral triangle, the triangle ABC is an isosceles triangle, ad = AE, ∠ DAE = 80 °, bad = 15 ° to find the degrees of ∠ CAE and ∠ EDC
- 10. 1. Given △ ABC, we make △ abd and △ ace with AB and AC as the sides, and ad = AB, AC = AE, ∠ DAB = ∠ CAE, and connect DC with be, G and F as DC respectively
- 11. As shown in the figure, ad is the height on the hypotenuse of the right angle △ ABC, de ⊥ DF, and de and DF intersect AB, AC and E, f respectively AD＝BE BD．
- 12. As shown in the figure, the chord AB and radius OC of ⊙ o are extended and intersected at point D, BD = OA. If ∠ AOC = 105 °, then ∠ D=______ Degree
- 13. As shown in the figure, the image of the quadratic function y = ax2-4x + C passes through the coordinate origin and intersects with the X axis at point a (- 4,0) (1) Find the analytic formula of quadratic function; (2) If there is a point P on a parabola, s △ AOP = 8, please write out the coordinates of point P directly
- 14. As shown in the figure, △ ABC, ab = AC, points D and E are on the extension line of AB and AC respectively, and BD = CE, de and BC intersect at point F. verification: DF = EF
- 15. As shown in the figure, in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at D, e is a point on AB, AF ⊥ CE is at F, ad intersects CE at G point. It is proved that ∠ B = ∠ CFD
- 16. As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = AC, points D and E are on BC and AC respectively, and BD = CE, M is the midpoint of AB, and △ MDE is isosceles triangle As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = AC, points D and E are respectively on BC and AC, and BD = CE, M is the midpoint of AB, and △ MDE is isosceles right triangle. Please explain the reasons
- 17. As shown in the figure, in the triangle ABC, BC = 6, e and F are the midpoint of AB and AC respectively, the moving point P is on the ray EF, BP intersects CE at D, bisector of angle CBP intersects CE at Q, when CQ = 1 / 3 * ce, EP + BP=____ .
- 18. As shown in the figure, in the right angle trapezoid ABCD, ad ‖ BC, ∠ ABC = 90 °, BD ⊥ DC, BD = DC, CE bisection ∠ BCD, crossing AB at point E, crossing BD at point h, en ∥ DC intersecting BD at point n ①BH=DH；②CH=( 2+1)EH；③S△ENH S△EBH＝EH EC. The correct one is () A. ①②③ B. Only ② and ③ C. Only two D. Only 3
- 19. As shown in the figure, in the isosceles △ ABC, ab = AC. the semicircle with diameter AB intersects BC at point D and AC at point E De = 40 °, find ∠ A and Degree of AE
- 20. Let d be a point on the edge BC of △ ABC, and AB2 = ad2 + BD × DC