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As shown in the graph, points D, e and F are the points on the edges BC, Ca and ab of the triangle ABC, De is parallel to Ba, DF is parallel to Ca, and it is proved that ∠ FDE = ∠ a
De parallel Ba, DF parallel CA
So the quadrilateral afde is a parallelogram
So ∠ FDE = ∠ a
I hope my answer can help you!
In △ ABC, the midpoint F of AB is dB, perpendicular to BC, perpendicular to e, and the extension line of intersection CA is at point D. if EF = 3, be = 4, angle c = 45 °, find the length of DF emergency
Make am ⊥ BC, the foot of foot is m, and ⊥ FBE is similar to ⊥ ABM, and the ratio of corresponding sides is 1 / 2, then the length of EM and am can be obtained. Because the angle c = 45 ° and △ AMC and △ Dec are right triangle, am = MC, de = EC = cm + MC, and then DF = de-ef, DF = 7
It is known that: as shown in the figure, △ ABC, the midpoint f passing through AB is de ⊥ BC, the perpendicular foot is e, and the extension line of intersection CA is at point D. if EF = 3, be = 4, ﹤ C = 45 °, the value of DF: Fe is______ .
AG ⊥ BC is used for crossing point a, and G is for perpendicular foot,
∵DE⊥BC∴EF∥AG
And the midpoint is ab
Ψ e is also the midpoint of BG, EF
AG=BF
AB=1
Two
∴EG=BE=4 AG=2EF=6
And ? C = 45 ° Ag = GC = 6
∴EC=EG+GC=10
And ∵ C = 45 ° de ⊥ BC
∴DE=EC=10
∴DF=DE-EF=10-3=7
∴DF：FE=7：3．
So the answer is: 7:3
In the triangle ABC, ab = AC, EF intersects AB with E, BC with D, extension of AC at F, and be = CF
Make the segment eh ‖ BC and intersect BC at H point
∵AB=AC,∠B=∠ACB =∠AEH=∠AHE,
∴AE=AH
/ / be = HC according to be = CF
∴HC=CF
EH ∵ DC
∴ED/DF=HC/CF
So de = DF
As shown in the figure, ∠ C = 90 ° CA = CB in △ ABC, point D is the midpoint of AB edge, e and F are respectively on Ca and CB, and ∠ EDF = 90 ° A. verification: de = DF Asking for reasons
It is proved that because ∠ C = 90 ° CA = CB, point D is the midpoint of AB side, so ∠ ACD = ∠ B, CD ⊥ AB, BD = AB / 2 = CD, (three lines in one) because ∠ EDF = 90 ° so ∠ EDC + ∠ CDF = 90 (vertical meaning) because ∠ CDF + ∠ BDF = 90, so ∠ EDC = ∠ FDB (the remainder of the same angle is equal) so △ EDC ≌
As shown in the figure RT △ ABC, ∠ ACB = 90 ° D is the midpoint De of AB, DF intersects AC to e and BC to f respectively, and de ⊥ DF, CA ⊥ CB It is proved that AE 2 + BF 2 = EF 2
It is proved that the point G of FD is extended so that DG = DF
∵ D is the midpoint of ab
∴AD＝BD
∵DE＝DF,∠ADG＝∠BDF
All △ ADG is equal to △ BDF (SAS)
∴AG＝BF,∠GAD＝∠B
∵∠ACB＝90
∴∠CAB+∠B＝90
∴∠CAB+∠GAD＝90
∴∠CAG＝90
∴AE²+AG²=EG²
∴AE²+BF²=EG²
∵DE⊥DF,DF＝DG
ν de vertical bisection FG
∴EF＝EG
∴AE²+BF²=EF²
In △ ABC, D is the midpoint of AB, extending Ca and CB to points E and f respectively, so that de = DF; passing through E and F, making the vertical lines of Ca and CB, intersecting at p. verification: PAE = PBF
As shown in the figure, take the midpoint m and N of AP and BP respectively, and connect em, DM, FN and DN
According to the triangle median line theorem, we can get: DM ‖ BP, DM = 1
2BP=BN，DN∥AP，DN=1
2AP=AM，
∴∠AMD=∠APB=∠BND，
∵ m and N are the middle points of AEP and BFP respectively,
∴EM=AM=DN，FN=BN=DM，
It is known that de = DF,
∴△DEM≌△FDN（SSS），
∴∠EMD=∠FND，
∴∠AME=∠BNF，
△ ame and △ BNF are isosceles triangles with equal vertex angles,
∴∠PAE=∠PBF．
In △ ABC, D is the midpoint of AB, extending Ca and CB to points E and f respectively, so that de = DF; passing through E and F, making the vertical lines of Ca and CB, intersecting at p. verification: PAE = PBF
As shown in the figure, take the midpoint m and N of AP and BP respectively, and connect em, DM, FN and DN. According to the triangle median line theorem, we can get: DM ∥ BP, DM = 12bp = BN, DN ∥ AP, DN = 12ap = am, amd = ∠ APB = ∠ Bnd, ∵ m, n are the midpoint of AEP and BFP hypotenuse of right triangle,
As shown in the figure, points a, B and C are on ⊙ o, and ab = AC, and the chord AE intersects BC with D. It is proved that ab ⊙ AE = ad · AE
∵AB=AC∴∠ACB=∠ABC=∠AEB
So △ abd is similar to △ AEB
AB / ad = AE / ab
AB 2 = ad · AE
AB is the diameter of ⊙ o, AC is the chord, CD ⊥ AB is in D. if AE = AC, be intersects ⊙ o at point F, connect CF, De, and prove AE ⊥ a = ad * ab
It is proved that AB is ⊙ o diameter, AC is chord, CD ⊥ AB in D = > right triangle ABC ∽ right triangle ACD
=>AB / AC = AC / ad = > AC? = ad * AB because AC = AE = > AE? = ad * AB
RELATED INFORMATIONS
- 1. As shown in the figure, the radius of ⊙ o is known to be 1, De is the diameter of ⊙ o, the tangent of ⊙ o is ad, C is the midpoint of AD, AE intersects ⊙ o at point B, and the quadrilateral bcoe is a parallelogram (1) Find the length of AD; (2) Is BC tangent to ⊙ o? If so, give proof; if not, give reasons
- 2. 0
- 3. AB is the diameter of the circle, the radius OC ⊥ AB, the chord EF is parallel AB through the midpoint D of OC. It is proved that the angle Abe is equal to 15 ° and the angle CBE is equal to 30 °
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- 5. Two equal line segments AB and CD coincide in one third. M and N are the midpoint of AB and CD respectively. If Mn = 12cm, find the length of ab
- 6. As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 120 ° and EF is the vertical bisector of ab. EF intersects BC at point F and ab at point E. verification: BF = 1 2FC．
- 7. It is known that in △ ABC, CA = CB, ∠ C = 90 ° D is any point on AB, AE ⊥ CD, perpendicular foot is e, BF ⊥ CD, and vertical foot is F. it is proved that EF = | ae-bf |
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- 9. As shown in FIG. 12, AOB is a straight line, OE bisection of AOC, of bisection of BOD, ∠ cod = 30 ° and try to find the degree of ∠ EOF
- 10. The surface area of regular hexagon is 12 If it is divided into six regular triangles, Is an equilateral triangle?
- 11. As shown in the figure, it is known that AB is parallel to CD, OA = OD, BC passes through point O, e, f are on the straight line AOD and AE = DF
- 12. As shown in the figure, AB is the chord of ⊙ o, points c and D are on the chord AB, and OC = OD
- 13. It is known that AB is the diameter of ⊙ O. the straight line BC and ⊙ o are tangent to point B. through a, ad ⊙ OC intersects ⊙ o at point D and connects CD Proof: CD is tangent of ⊙ o
- 14. It is known that AC and BD are two mutually perpendicular chords of the circle O: x2 + y2 = 4, and the perpendicular foot is m (1, 2) The maximum area of the quadrilateral ABCD is () A. 4 B. 4 Two C. 5 D. 5 Two
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- 17. As shown in the figure, in △ abd and △ ace, ab = ad, AC = AE, ∠ bad = ∠ CAE, connecting BC and de intersecting at point F, BC and ad intersecting at point g. verification: BC = De
- 18. In the triangle ABC, the points D and E are on BC, the angle CAE is equal to angle B, e is the midpoint of DC, BD is equal to AC, and the ad bisection angle BAE (1) when the angle BAC is equal to 90 degrees In the triangle ABC, points D and E are on BC, the angle CAE is equal to angle B, e is the midpoint of DC, BD is equal to AC, and ad bisects the angle BAE (1) When the angle BAC is equal to 90 degrees, it is proved that BD is equal to AC (2) When the angle BAC is not equal to 90 degrees, is there still BD equal to AC? Explain the reason
- 19. 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
- 20. 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