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As shown in the figure, ∠ ABC = ∠ DBE = 90 °, DB = be, ab = BC (1) Confirmation: ad = CE, ad ⊥ CE; (2) If the external conditions of △ DBE rotating around point B to △ ABC remain unchanged, then the conclusion in (1) still holds? Draw a graph to prove your conclusion
(1) In △ abd and △ CBE, ab = BC, abd = cbebd = be, Δ abd ≌ △ CBE (SAS), ∵ ad = CE, bad = ≌ BCE. ? AGB and ≌△ CBE (SAS), ? ad = CE, bad = ∵ BCE
As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, AC = BC, D is the midpoint of BC edge, CE ⊥ ad intersects point E, BF ∥ AC, and the extension of CE is at point F Confirmation: BD = BF
It is proved that in ∵ RT △ ABC, ∵ ACB = 90 °, AC = BC,
∴∠1+∠2=90°，
∵BF∥AC，
∴∠ACB=∠CBF=90°，
∵CE⊥AD，
∴∠2+∠3=90°，
∴∠1=∠3，
In △ ACD and △ CBF,
A kind of
∠1＝∠3
AC＝BC
∠ACB＝∠CBF ，
∴△ACD≌△CBF，
∴BF=CD，
∵ D is the midpoint on the edge of BC,
∴BD=CD，
∴BD=BF．
As shown in the figure, ad is the center line of △ ABC, CE ⊥ ad is in E, BF ⊥ ad, and the extension line of ad is at F
It is proved that ∵ ad is the center line on the BC side of △ ABC,
∴BD=CD．
∵ CE ⊥ ad in E, BF ⊥ ad,
∴∠BFD=∠CED．
In △ BFD and △ CED
∠F＝∠CED
∠BDF＝∠CDE
BD＝CD ，
∴△BFD≌△CED（AAS）．
∴CE=BF．
As shown in the figure, it is known that in RT △ ABC, AC = BC, D is the midpoint of BC, CE ⊥ ad is at e, BF ∥ AC intersects with CE extension line at point F. verification: AC = 2BF
Because ∠ ACB = 90 ° and CE ⊥ ad, ∠ CED = 90 ⊥
So right triangle ACD is similar to right triangle CDE
So ∠ CAD = ∠ DCE
Because BF ‖ AC, ∠ CBF = 90 °
So right triangle ACD and right triangle CBF are congruent
So CD / AC = BF / BC = BF / AC
So BF = CD = 1 / 2BC = 1 / 2Ac, so AC = 2BF
As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, CD ⊥ AB, the perpendicular foot is D, e and F are the points on the edge of AC and BC respectively, and CE = 1 / 3aC, BF = 1 / 3bC 1) Prove AC / BC = CD / BD 2) Find the degree of ∠ EDF
First question: because: ACB = 90 ° and CD ⊥ AB, so: CD * CD = ad * BD (two triangles know each other or what law) so: AC / BC = CD / BD the second problem: CE = 1 / 3aC, BF = 1 / 3bC, so: CE / BF = AC / BC and the first question: AC / BC = CD / bdce / BF = CD / BD 1
As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, AC = BC, D is the midpoint of BC edge, CE ⊥ ad intersects point E, BF ∥ AC, and the extension of CE is at point F Confirmation: BD = BF
In the case of CBF, it is determined that the CBF is ∵ CBF \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\bfd = CD
In the RT triangle ABC, the angle ACB = 90 degrees, AC = BC, D is the midpoint of BC edge, CE is perpendicular to ad, the perpendicular foot is e, BF is parallel to AC, and the extension line of intersection CE is at point F If AC = 12, find the length of DF
It is proved that because BF ∥ AC, so ﹣ ACD + ∠ CBF = 180 °∠ ACF = ∠ BFC, and because ∠ ACB = 90 ° and ∠ CEA = 90 ° so ∠ CBF = 90 °∠ CAD + ∠ ACF = ∠ CAD + ∠ ADC = 90 ° i.e., ∠ ACF = ∠ ADC, so ∠ BFC = ∠ ADC in △ ADC and △ CFB, AC = CB ∠ ACB = ∠ CBF = 90 °
As shown in the figure, in the RT triangle ABC, the angle ABC = 90 degrees, AC = BC, D is the midpoint of BC, CE is vertical ad, the perpendicular foot is point E, BF / / AC, the extension line of CE is at point F. if AC = 12, find DF
If the angle B = 90, then AC is the hypotenuse. How is the hypotenuse equal to the right angle?
Is it a mistake
It should be angle ACB = 90
So it's an isosceles right triangle
And DB = 12 / 2 = 6
In RT triangle DBF, only BF is required
It can be proved that triangular CBF is equal to triangular ACD
AC=CB=12
Angle ACD = angle CBF = 90
Angle FCD + angle ADC = angle CAD + angle ADC = 180-90
So angle FCD = angle CAD
In this way, triangular CBF is equal to triangular ACD
So FB = CD = 6
FD²=BD²+FB²=72
DF = 6 root number 2
As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, AC = BC, D is the midpoint of BC edge, CE ⊥ ad intersects point E, BF ∥ AC, and the extension of CE is at point F Confirmation: BD = BF
In the case of CBF, it is determined that the CBF is ∵ CBF \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\bfd = CD
It is known as follows: in RT △ ABC, ∠ C = 90 ° and fold the triangle along a line be passing through point B, so that point C coincides with point D on edge ab. when ∠ A is what, point D is exactly the midpoint of AB? Write an angle that you think is appropriate and use the size of this angle to prove that D is the midpoint of ab
When ∠ a = 30 °, point D is exactly the midpoint of ab. (2 points)
It is proved that: ∵ a = 30 °, C = 90 °,
∴∠CBA=60°．
And △ BEC ≌ △ bed,
﹤ CBE = ∠ DBE = 30 ° and ﹣ EDB = ∠ C = 90 ° and ∠ EBA = ∠ a,
ν be = AE, and ∠ EDB = 90 °, that is ed ⊥ ab
﹤ D is the midpoint of ab. (6 points)
RELATED INFORMATIONS
- 1. This is the geometry of grade two, which is related to parallelogram In the RT triangle ABC, the angle c = 90 degrees, M is the midpoint of AB, am = an, Mn is parallel to AC, and Mn = AC
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- 3. It is known that the triangle ABC, CA = CB, point O is on the vertical bisector of Ca and CB, and m.n is on a straight line AC.BC Go ahead, It is proved that CN + Mn = am
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- 6. As shown in the figure, both the triangle ABC and the triangle Dec are isosceles right triangles. The angle ACB = angle DCE = 90 ° f is the midpoint of De, h is the midpoint of AE, and G is the midpoint of BD If the △ Dec in Fig. 1 is rotated clockwise by an acute angle around point C to Fig. 2, is the triangle FGH still an isosceles right triangle
- 7. As shown in the figure, in diamond ABCD, AE ⊥ BC, perpendicular foot is e, angle DAE = 2, angle BAE, BD = 15, find the length of AC, ab
- 8. As shown in the figure, in △ ABC, D is a point on AB, and ad = AC, AE ⊥ CD, the perpendicular foot is e, and F is the midpoint of CB
- 9. As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ BC, the bisector of ⊥ BAC is AE at point E, EF ⊥ AB is at point F, eg ⊥ AC intersects AC extension at point g. verification: BF = CG
- 10. As shown in the figure, in the triangle ABC, ad is the bisector of ∠ BAC, EF bisects ad vertically, intersects ad at e, and the extension line intersects BC at F, then ∠ B = ∠ caf? Because ad bisects angle BAC, angle bad = angle CAD Because EF bisects ad vertically, AF = DF, so angle DAF = angle ADF Because angle DAF = angle DAC + angle CAF, angle ADF = angle B + angle bad, angle DAC + angle CAF = angle B + angle bad Because angle bad = angle CAD, angle B = angle caf Why angle ADF = angle B + angle bad
- 11. In the triangle ABC, BD = DC = AC, D is the midpoint of BC and E is the midpoint of DC. It is proved that the bisection angle BAE of AD
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- 13. As shown in the figure, in the isosceles triangle ABC, ab = AC, D is a point on the extension line of AB, BD = AB, CE is the center line on waist AB, and verify; CD = 2ce needs specific process
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- 16. As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained
- 17. The graph of the known quadratic function y = AX2 + BX + C (a ≠ 0) is shown in the figure ① ABC > 0; ② B < A + C; ③ 4A + 2B + C > 0; ④ 2C < 3B; ⑤ a + b > m (am + b) (m ≠ 1 real number) The correct conclusion is () A. 2 B. Three C. Four D. Five
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- 19. As shown in the figure: in isosceles RT △ ABC, ∠ C = 90 degrees, AC = 8, f is the midpoint of AB edge, points D and e move on AC and BC sides respectively, and keep ad = CE, connect We need the whole proof process. Thank you
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