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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
It is proved that DN ⊥ AC, DM ⊥ AB are made by D,
△ the area of DBF is: 1
2BF•DM，
The area of △ DCE is 1
2DN•CE，
The area of ∵ △ DCE and △ DBF are equal,
∴1
2BF•DM=1
2DN•CE，
∵CE=BF，
∴DM=DN，
Ψ ad bisection ∠ BAC (points with equal distance to both sides of the corner are on the bisector of the angle)
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
It is proved that DN ⊥ AC, DM ⊥ AB are made by D,
△ the area of DBF is: 1
2BF•DM，
The area of △ DCE is 1
2DN•CE，
The area of ∵ △ DCE and △ DBF are equal,
∴1
2BF•DM=1
2DN•CE，
∵CE=BF，
∴DM=DN，
Ψ ad bisection ∠ BAC (points with equal distance to both sides of the corner are on the bisector of the angle)
As shown in the figure, ad is the angular bisector of the triangle ABC, EF is two points on AC and ab respectively, CE = BF, prove; s △ DCE = s △ DBF
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Inside angle of triangle ABC and 180 degrees
Angle a = 40 degrees, angle B = 60 degrees, then angle ACB = 180-40-60 = 80 degrees
CE is high in the right angle delta AEC
Angle a = 40 degrees, ∠ AEC = 90, then ∠ ace = 180-40-90 = 50 degrees
If CD is an angular bisector, then ∠ ACD = 40
∠ACE=∠ACD+∠DCE
∠ DCE = 50-40 = 10 degrees
As shown in the figure, it is known that RT △ ABC ≌ RT △ Dec, ≌ e = 30 ° and D is the midpoint of AB, AC = 1. If △ Dec rotates clockwise around point d so that ED, CD and right angle sides BC of RT △ ABC intersect with m, n respectively, then when △ DMN is an equilateral triangle, the value of AM is () A. Three B. 2 Three Three C. Three Three D. 1
In RT △ ABC, ∠ e = 30 ° and D is the middle point of AB, then △ BCD, BC = 3, ∠ CDB = 120 °, CD = BD, passing point D as DP ⊥ BC at point P, then PC = 32, DP = PC · Tan 60 ° = 12. In RT △ DMP, MP = DP · tan30 ° = 36,
As shown in the figure, ∠ ABC = 90 ° in RT △ ABC. Rotate RT △ ABC clockwise for 60 ° around point C to obtain △ Dec, point E on AC, and then flip RT △ ABC along the line of AB by 180 ° to obtain △ ABF. Connect ad （1） The results show that the quadrilateral AFCD is rhombic; (2) Connect be and extend intersection ad to g, connect CG, what is quadrilateral ABCG special parallelogram and why?
(1) It is proved that RT △ Dec is obtained by the rotation of RT △ ABC around point C by 60 ° and AC = AF, ABF = 90 ° and ∵ AC = AF, ABF = 90 ° and ∵ AC = AF, respectively
As shown in the figure, ∠ ABC = 90 ° in RT △ ABC. Rotate RT △ ABC clockwise for 60 ° around point C to obtain △ Dec, point E on AC, and then flip RT △ ABC along the line of AB by 180 ° to obtain △ ABF. Connect ad （1） The results show that the quadrilateral AFCD is rhombic; (2) Connect be and extend intersection ad to g, connect CG, what is quadrilateral ABCG special parallelogram and why?
(1) It is proved that RT △ Dec is obtained by the rotation of RT △ ABC around point C by 60 ° and AC = AF, ABF = 90 ° and ∵ AC = AF, ABF = 90 ° and ∵ AC = AF, respectively
As shown in the figure, in the RT triangle ABC, the angle ABC is equal to 90 degrees. Rotate the RT triangle ABC clockwise about point C by 60 degrees The triangle Dec is obtained, the point E is on AC, and then the RT triangle is turned 180 degrees along the line to get the triangle ABF connection ad. it is proved that the quadrilateral AFCD is a diamond Urgent!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
∵ E on AC, ACB = 60
∵ triangle ABC and △ ABF are symmetric for ab
∴AF=AC,CF=2CB=AC=AF
AC = DC in △ ACD, ∠ ACD = 60,
The △ ACD is an equilateral triangle
∵∠FAC=∠DAC=60
So AF is parallel to DC
Similarly, it can be proved that DA and CF are parallel and equal
So the quadrilateral AFCD is a diamond
It is known that △ ABC and △ Dec are equilateral triangles, ∠ ACB = ∠ DCE = 60 ° and B, C and E are on the same straight line. Connect BD and AE. Find the degree of ∠ AHB
BD and AE are equilateral triangles ? ABC and △ Dec are equilateral triangles BC = AC, CD = CE, ∠ BCA = ∠ DCE = 60 °∠ BCD = ∠ ace = 60 ° + ∠ ACD BCD ? CBD = ? CAE ? bah - ? ABH = 180 ° - (? BAC + ∠ CAE) - (∠ ABC - ∠
As shown in the figure, △ ABC and △ Dec are equilateral triangles, b.e.c is on a straight line, AE and BD intersect at point h, AC and BD intersect at point P, AE and CD intersect at point Q. it is proved that PQ is parallel to be
Because positive △ ABC, positive △ Dec
Therefore: BC = AC, CD = CE, ∠ ACB = ∠ DCE = 60 °
Because b.e.c. is in a straight line
Therefore: ∠ ACD = 60 °
Therefore: ∠ BCD = ∠ ace = 120 °
Therefore: △ BCD ≌ △ ACE (SAS)
Therefore: ∠ QAC = ∠ PBC (combined with ∠ ACD = ∠ ACB = 60 ° BC = AC)
Therefore: △ BCP ≌ △ ACQ (ASA)
Therefore, PC = QC
Therefore: △ PCQ is a regular triangle
Therefore: ∠ QPC = 60 ° = ∠ ACB
Therefore: PQ ∥ be
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. 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
- 7. 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
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- 9. In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE
- 10. 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
- 11. 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
- 12. 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
- 13. 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
- 14. 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
- 15. 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
- 16. 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
- 17. 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
- 18. 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
- 19. As shown in the figure, in RT △ ABC, ab = AC, ∠ a = 90 °, point D is any point of BC, DF ⊥ AB is in F, de ⊥ AC is in E, M is the midpoint of BC. Try to judge what shape △ MEF is and prove your conclusion
- 20. D is the middle of the edge BC of the triangle ABC, De is vertical AC, DF is vertical AB, the vertical feet are e, F, and BF = CE, judge the shape of triangle ABC Is it an isosceles right triangle? Give proof