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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
∵ a quadrilateral ABCD is a rhomboid ∵ ad ∵ BC ∵ AE ⊥ BC ∵ DAE = 90 BAE = 45 ° when DAB is an acute angle, then ∠ DAB = ∠ Dae - ∠ BAE = 45 °: point E is on the extension line of CB, passing through point B is BF ⊥ ad at F, there is ⊥ AFB is isosceles right triangle, let AB = a, there is AF = BF = √ 2 / 2a in right triangle
As shown in the figure, there is a point E on the BC side of the square ABCD, and the bisector of DAE intersects CD at F. try to explain AE = DF + be by using the method of rotation
The result shows that: (1) if the angle of ∞ is ∞, then ∞ aff = ????????????
As shown in the figure, there is a point E on the edge BC of the square ABCD. The intersection CD and F of the bisector are obtained. It is proved that AE = DF + be
It is proved that: take a point G on the extension line of CB, make BG = DF, connect Ag ∵ square, ABCD ∵ ab ∵ AB = ad, ∵ ABG ∵ ADC = 90 ? BG ≌ BG = DF ≌ ADF (SAS) ? g ? g ? g, make BG = DF, make BG = DF, connect Ag ∠ BAE ∵ ab ∵ C
As shown in the figure, in the square ABCD, F is the midpoint of CD, E is a point on the edge of BC, and AF bisects [DAE], to verify: AE=EC+CD
It is proved that ∵ AF bisection ∵ DAE, ∵ d = 90 ° FH ⊥ AE,
∴∠DAF=∠EAF，FH=FD，
And ∵ DF = FC = FH, Fe is the common edge,
∴△FHE≌△FCE（HL）．
∴HE=CE．
∵AE=AH+HE，AH=AD=CD，HE=CE，
∴AE=EC+CD．
In the square ABCD, e is a point on BC, AF bisects ∠ DAE and intersects CD with F. it is proved that AE = be + DF
From ab = ad, BG = DF, ABG = ADF = ADF = ADF ∵ ABCD is a square, ᙽ AB = ad, ᚛ bad ∵ bad ∵ ABC ∵ D ∵ d = 90 ∵ a ∵ ABC = 90 ∵ ABG = 90 ° from ab = ad, BG = DF, ABG = ADF = 90 °, from ab = ad, BG = DF, ABG = ADF = ADF, ≌ g ≌ ADF, ? g ? AFD ? bag ᙨ bag ᙨ DAF g = B
It is proved that AE = be + DF
Rotate △ Abe with point a as the origin, so that ab coincides with AD, e to e '
AE=AE' ,BE+DF=E'F
E'F‖AB,
∴∠AFE'=∠BAF=∠BAE+∠EAF=∠DAF+∠DAE'=∠E'AF
So AE '= e'f
That is, AE = be + DF
As shown in the figure, ab = AC, ad bisection angle BAC, AB bisection angle DAE, AE vertical be, vertical foot E
0
0
The intersection point of AM and CD is point n. extend am to f to make MF = am
∵BM=EM
﹤ Abe is a parallelogram
∴BF=AE ∠ABF+∠BAE=180°
∵∠BAC=∠DAE=90°
∴∠CAD+∠BAE=180°
∴∠ABF=∠CAD
∵BF=AE AD=AE
∴BF=AD
∵AB=AC
∴△ABF ∽△CAD
∴∠BAF=∠ACD
∵∠BAC=90°
∴∠BAF+∠CAN=90°
∴∠ACD+∠CAN=90°
∴∠ANC=90°
∴AM⊥CD
3 as shown in the figure, ab = AC, ad = AE, M is the midpoint of be, ∠ BAC = ∠ DAE = 90 °. Verification: am ⊥ DC
It is proved that: as shown in the figure, the plane rectangular coordinate system is established with point a as the origin, AC as the positive half axis of X axis and ab as the negative half axis of Y axis. Let e (m, n), C (R, 0), then: D (n, - M), B (0, - R), m (M / 2, (N-R) / 2)
AB equals AC, ad equals AE, angle BAC equals angle, DAE equals 90 degrees, M is the midpoint of be, find am vertical DC, thank you Thank you for your trouble. The common vertex of ade and ADC is a
Drawing shows that the quadrilateral BCDE is isosceles trapezoid
Extended Ma cross DC and F
∠CBA+∠CDA=90
∠CBE+∠CDE=180
∠ABM+∠ADF=90
Therefore, ABM + ADF = 90
M is the midpoint of be, then am = MB
∠MBA=∠MAB=∠DAF
Therefore, ADF + DAF = 90
So am vertical DC
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. 0
- 7. In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE
- 8. 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
- 9. 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
- 10. As shown in the figure, AB is the chord of ⊙ o, points c and D are on the chord AB, and OC = OD
- 11. 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
- 12. As shown in the figure, D, e and F are points on the three sides of △ ABC, and the areas of CE = BF, △ DCE and △ DBF are equal Verification: ad bisection ∠ BAC
- 13. In △ ABC, ∠ a = 1 2∠B=1 3 ∠ ACB, CD is the height of △ ABC, CE is the angular bisector of ∠ ACB, and the degree of ∠ DCE is calculated
- 14. 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
- 15. As shown in the figure, in the RT triangle ABC, the angle c = 90 degrees, the DB bisector angle ABC intersects AC at point D, De is the vertical bisector of AB and intersects AB at point E (1) Find the degree of angle A (2) If BC = 6, AC = 8, then the circumference of the triangle BDC is
- 16. 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
- 17. 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
- 18. 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
- 19. In △ ABC, ab = BC, ∠ B = 90 °, M is the midpoint of AC, D.E is the moving point on AB and BC respectively, and BD = CE, the shape of △ DEM is obtained and proved
- 20. 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