![](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 △ ABC, ab = AC point D is a point on CB extension line ∠ ADB = 60 ° point E is a point on AD and de = dB, it is proved that AE = be + BC Pictures can be drawn according to the theme!
Make an auxiliary line AG, Ag ⊥ BC
∵AB=AC
ν Ag vertically bisects BC
And ∵ ADB = 60 ° and de = dB
∴∠DAG=30°,DE=DB=BE
DB+BG=1/2(DE+AE)
DB + 1 / 2BC = 1 / 2 (de + AE) both sides are multiplied by 2
2DB+BC=DE+AE
AE=2DB+BC-DE=DB+BC=BE+BC
The proof is complete
As shown in the figure, it is known that in △ ABC, ab = AC, D is a point on CB extension line, ∠ ADB = 60 ° e is a point on ad, and there is de = dB. Verification: AE = be + BC
It is proved that: extend DC to F, make CF = BD, connect AF, ∵ AB = AC, AB = AC, ABC = ∠ ACB, abd = ∠ ACF. In △ abd and △ ACF, ab = AC ∠ abd = CF, △ abd ≌ △ ACF (SAS), ad = AF, and ? ADB = 60 °
As shown in the figure, in △ ABC, ∠ ACB = 90 ° AC = BC, AE is the center line of BC, passing point C is CF ⊥ AE at F, passing through B is BD ⊥ CB, and the extension line of CF is at point D (1) Confirmation: AE = CD; (2) If BD = 5cm, find the length of AC
(1) Proof: DB ⊥ BC, CF ⊥ AE,
∴∠DCB+∠D=∠DCB+∠AEC=90°．
∴∠D=∠AEC．
In △ DBC and △ ECA,
∠D＝∠AEC
∠ACE＝∠CBD
AC＝CB
∴△DBC≌△ECA（AAS），
∴AE=CD；
（2）∵△DBC≌△ECA，
∴BD=CE，
∵ AE is the center line on the edge of BC,
∴BC=2CE=2BD=10cm，
∴AC=BC=10cm．
In △ ABC, ab = CB, ∠ ABC = 90 ° f is a point on the extension line of AB, point E is on BC, and AE = CF (1) It was proved that be = BF; (2) If ∠ CAE = 30 °, calculate the degree of ∠ ACF
(1) It is proved that as shown in the figure, ∵ ABC = 90 °, in RT △ Abe and RT △ CBF
AB＝CB
CF＝AE ，
∴Rt△ABE≌Rt△CBF（HL），
∴BE=BF；
（2）∵AB=CB，∠ABC=90°，
∴∠BAC=∠BCA=45°，
∵∠CAE=30°，
∴∠BAE=45°-30°=15°，
∵Rt△ABE≌Rt△CBF，
∴∠BCF=∠BAE=15°，
∴∠ACF=∠BCF+∠BCA=15°+45°=60°．
AB = CB, ∠ ABC = 90 °, f is a point on AB extension line, point E is on BC, and AE = CF, RT triangle Abe is all equal to rtcbf
prove:
∵∠ABC＝90
∴∠FBC＝180-∠ABC＝90
∴∠ABC＝∠FBC
∵AB＝CB、AE＝CF
∴Rt△ABE≌Rt△CBF （HL）
It is known that △ ABC is an equilateral triangle, passing through point D on the edge of AC as DG ∥ BC, crossing AB with point G, taking point E on the extension line of GD to make de = DC, connecting AE, BD 1. Find that △ age is equal to DAB 2. Make ef ‖ DB through point E, intersect BC with point F, connect AF, and calculate the degree of ∠ AFE
It is proved that: if DG ≌ BC, then ≌≌≌ EGA (SAS). 2. If ⊿ ADG is an equiltriangle triangle. If ⊿ ADG is an equiltriangle. If ⊿ Ag = ad = ad = DG; if ⊙ bad = EGA = 60.de = DC, then de + DG = DC + ad, that is, eg = AC = AC = ab. therefore ⊿ bad ≌ EGA (SAS). 2 ? bad ? EGA (has been proved), then: 57; abd = ≌ BD = AE; BD = AE; EF ? EF ≌ EF ≌ EF BD, DG
It is known that, as shown in the figure, △ ABC is an equilateral triangle, passing through point D on the edge of AC as DG ‖ BC, crossing AB with point G, taking point E on the extension line of Gd, making de = DC, connecting AE and BD (1) Verification: △ age ≌ △ DAB; (2) Let ‖ e be the intersection point of AF
(1) It is proved that: ? ABC is an equilateral triangle, DG ∥ BC, AGD = ∠ ABC = 60 °, ADG = ∠ ACB = 60 ° and ∠ BAC = 60 °, △ AGD is an equilateral triangle, Ag = GD = ad, ∠ AGD = 60 °. ? de = DC, ? Ge = GD + de = AD + DC = AC = AB, ? in △ age and △ DAB, Ge = ab ∠ AGD
It is known that in △ ABC, ∠ C = 90 ° D is the midpoint of AB, e and F are on AC and BC respectively, and de ⊥ DF. Verification: AE2 + BF2 = ef2
It is proved that am ∥ BC is made through point a, and FD extension line is crossed at point M,
Connect em
∵AM∥BC，
∴∠MAE=∠ACB=90°，∠MAD=∠B．
∵AD=BD，∠ADM=∠BDF，
∴△ADM≌△BDF．
∴AM=BF，MD=DF．
And ∵ de ⊥ DF, ∵ EF = em
∴AE2+BF2=AE2+AM2=EM2=EF2．
As shown in the figure, △ ABC, ab = AC, D is the midpoint of BC, the straight line passing through point a EF ‖ BC, and AE = AF, verification: de = DF
Proof: as shown in the figure, connect ad
In ABC, ab = AC, D is the midpoint of BC,
∴AD⊥BC，
∵EF∥BC，
∴AD⊥EF，
AE = AF,
ν ad vertical bisection EF,
∴DE=DF．
As shown in the figure, △ ABC, ab = AC, D is the midpoint of BC, e and F are the points on AB and AC respectively, and AE = AF, verification: de = DF
Proof: connect ad,
∵ AB = AC, D is the midpoint of BC,
∴∠EAD=∠FAD，
In △ AED and △ AFD,
AE＝AF
∠EAD＝∠FAD
AD＝AD ，
∴△AED≌△AFD（SAS），
∴DE=DF．
RELATED INFORMATIONS
- 1. As shown in the figure, in △ ABC, ab = AC, D is the midpoint of BC, points E and F are on AB and AC respectively, and AE = AF, it is proved that de = DF
- 2. As shown in the figure, translate the right triangle ABC along the BC direction to the position of the right triangle def, ab = 8, be = 5, Ge = 5
- 3. In the triangle ABC, the angle c = 90 degrees, D is a point on the edge AC, and ad = 1, angle abd = 45 degrees. Let the angle CBD = a, BC = X. when Tana = 3 / 5, find x =? When x = 0.16, find Tana =?
- 4. As shown in the figure, in △ ABC, ∠ ACB = 90 °, point E is the midpoint of AB, connecting CE, passing through point E as ed ⊥ BC at point D, taking a point F on the extension line of De to make AF = CE
- 5. As shown in the figure, in the triangle ABC, the vertical bisector L1 of the AB side In the triangle ABC, the vertical bisector L1 of AB side intersects BC at D, L2 of AC side intersects BC at e, Li and L2 intersect at point O. the perimeter of triangle ade is 6 and that of OBC of triangle is 16. Find the length of Ao
- 6. It is known that in the square ABCD, ∠ man = 45 ° and ∠ man rotate clockwise around point A. both sides of it intersect CB, DC (or their extension) at point m and N, ah ⊥ Mn at point H (1) As shown in Fig. 1, when ∠ man point a is rotated to BM = DN, please write down the quantitative relationship between ah and ab directly______ ; (2) As shown in Fig. 2, when ∠ man rotates around point a to BM ≠ DN, does the quantitative relationship between ah and AB in (1) still hold? If not, please write down the reason, if yes, please prove it; (3) As shown in Fig. 3, given ∠ man = 45 °, ah ⊥ Mn at point h, and MH = 2, NH = 3, find the length of ah
- 7. As shown in the figure, the turning point of a highway is an arc. The point O is the center of the arc, AB is equal to 120m, C is the point on AB, O is the point on AB, the perpendicular foot is D, and CD is equal to 20m. Calculate the radius of the curve
- 8. It is known that in △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of AC, AE ⊥ BD is at e, and extended AE is crossed with BC at F. it is proved that ∠ ADB = ∠ CDF
- 9. As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = 4, AC = 4, now translate △ ABC along CB direction to △ a ′ B ′ C ', if the translation distance is 3 (1) Calculate the area of the overlapped part of △ ABC and △ a ′ B ′ C '; (2) If the translation distance is x (0 ≤ x ≤ 4), find the area y of the overlapping part of △ ABC and △ a ′ B ′ C ', then what is the relationship between Y and X
- 10. It is known that ad is an angular bisector of △ ABC, where de ‖ AC intersects AB at point E and DF ‖ AB intersects AC at point F Verification: quadrilateral AEDF is rhombic
- 11. As shown in the figure, in the RT triangle ABC, ∠ = 90 ° ad bisect ∠ BAC, and ∠ B = 3 ∠ bad, find the degree of ∠ ADC
- 12. As shown in the figure, ad is the center line of △ ABC, ∠ ADC = 45 ° and BC = 2cm. Fold △ ACD along ad so that point C falls at the position of E, then be=______ cm．
- 13. As shown in the figure, take the diameter BC of ⊙ o as an equilateral ⊙ ABC, AB and AC intersect ⊙ o at D and e. it is proved that BD = de = EC
- 14. It is known that in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is () A. 30° B. 36° C. 45° D. 50°
- 15. As shown in the figure, in △ ABC, ∠ B = 90 ° o is a point on AB, and a circle with o as its center and ob as its radius intersects with point e at point E
- 16. 0 is a point in the triangle ABC, and the distance from O to the three sides is equal. Given the angle a = 70 °, calculate the degree of BOC
- 17. It is known that: point I is the inner heart of the triangle ABC, point O is the outer center of the triangle ABC, ∠ BOC = 140 ° to find ∠ BIC? Why is ∠ a = 70 ° or 110 ° more detailed?
- 18. As shown in Figure 1, in triangle ABC, OB and OC are bisectors of angle ABC and angle ACB respectively. If angle a = x degrees, what is the degree of angle BOC
- 19. As shown in the figure, BC is the chord of circle O, OD ⊥ BC is at point E, intersection arc BC is at point D, and point a is a moving point on the major arc BMC (not coincident with B and C). BC = 4, radical 3, ed = 2 are known (1) Find the radius of circle O (2) Find the degree of ∠ a (3) Find the position of point a when the shadow part takes the maximum value I can't figure it out
- 20. In RT △ ABC, ∠ C = 90 ° cab = 60 ° and the distance between D and ab is 3.8cm, then how many centimeters is BC The properties of detailed points, especially the right triangle with 30 degree angle