![](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>)
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
Triangle AEB and triangle ADB, AB are common. EB is equal to ad, angle EBA is equal to angle, DAB is equal to 45 degrees. Congruence can be obtained from "edge, corner and edge"
As shown in the figure, in △ ABC, ad ⊥ BC is in D, ad ^ 2 = BD * DC. It is proved that ⊥ ABC is a right triangle
Using similar triangles
It is known that CD ⊥ AB is in D, AC = 3, BC = 4, ad = 9 / 5. (1) find the length of DC and BD; (2) prove that △ ABC is a right triangle
CD⊥AB
According to Pythagorean theorem
cd=12/5
bd=16/5
ad+bd=5
From the inverse theorem
3,4,5 are a group of Pythagorean numbers
Angle c = 90
Delta ABC is a right triangle
As shown in the figure, △ ABC and △ DBE are isosceles right triangles (1) Results: ad = CE; (2) Confirmation: AD and CE are vertical
It is proved that: (1) ∵ ABC and △ DBE are isosceles right triangles,
∴AB=BC，BD=BE，∠ABC=∠DBE=90°，
∴∠ABC-∠DBC=∠DBE-∠DBC，
Abe = CBD,
∴△ABD≌△CBE，
∴AD=CE．
(2) In prolonged ad, BC and CE were crossed to G and f respectively,
∵△ABD≌△CBE，
∴∠BAD=∠BCE，
∵∠BAD+∠ABC+∠BGA=∠BCE+∠AFC+∠CGF=180°，
And ∵ BGA = ∵ CGF,
∴∠AFC=∠ABC=90°，
∴AD⊥CE．
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, in RT △ ABC, the angle ABC = 90 ° AC = BC, D is the midpoint of BC, CE ⊥ ad, vertical is e, BF ∥ verification: AC As shown in the figure, we know that in RT △ ABC, the angle ABC = 90 ° AC = BC, D is the midpoint of BC, CE ⊥ ad, vertical is e, BF ∥ AC, the extension line of intersection CE is at point F, verification: AC = 2BF I can't send out the graph. Maybe the graph is the one to verify AB vertical bisection DF
AC = BC, D is the midpoint of BC. AC = 2CD. ? ACB = 90 °, BF ∥ AC. CBF = 90 °. ? CE ⊥ ad ᙽ CED = 90 °. In △ ACD and △ CED, ? CDA = ∠ CDE, ∠ ACD = ∠ CED, so ? ACD ∵ CED. ∵ ECD = ∠ EAD
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, 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,
As shown in the figure, in RT △ ABC, ∠ C = 90 °, a = 30 °, BD is the bisector of ∠ ABC, BC = 30, find the length of AD
In RT △ ABC, ∠ C = 90 ° and ∠ a = 30 ° so ∠ ABC = 60 °,
Because BD is the bisector of ABC, CBD = 30,
∴∠ABD=∠A=30°，
∴AD=BD，
In RT △ ABC,
∴cos30°=BC
BD，
∴BD=BC
cos30°=30
Three
2=20
3，
∴AD=20
3．
It is known as follows: in RT △ ABC, ∠ C = 90 ° and fold △ ABC along a line be passing through point B, so that point C falls exactly at the midpoint D of edge AB, then ∠ a=______ Degree
∵ in RT △ ABC, ∵ C = 90 ° and ᙽ BCE coincides with △ BDE,
∴ED⊥AB，∠EBA=∠EBC，
And point D is the midpoint of AB, and ﹤ AEB is an isosceles triangle,
∴∠A=∠EBA．
∵∠A+∠EBA+∠EBC=90°，
∴3∠A=90°，∴∠A=30°．
RELATED INFORMATIONS
- 1. As shown in the figure, in RT triangle ABC, angle c = 90 °, angle 1 = equal to angle 2, CD = 1.5, BD = 2.5?
- 2. 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
- 3. 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
- 4. 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
- 5. As shown in the figure, ab = AC, ad = AE are known
- 6. 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
- 7. 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
- 8. 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
- 9. As shown in the figure, in the triangle ABC, ad is the center line on the BC side. The circumference of the triangle abd is 5 smaller than that of the triangle ACD. Can you find the difference between the edge lengths of AC and ab?
- 10. As shown in the figure, we know AB = CD = 1, ∠ ABC = 90 ° and ∠ CBD = 30 ° to find the length of AC
- 11. 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
- 12. 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
- 13. 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．
- 14. 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
- 15. 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
- 16. 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
- 17. 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
- 18. 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
- 19. 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=____ .
- 20. 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