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As shown in the figure, in the triangle ABC, ∠ ACB = 90 °, ad bisection ∠ BAC, de ⊥ AB is verified in E; the straight line ad is the vertical bisection of CE, seeking the help of the great God
It is proved that: ∵ de ⊥ AB, ∵ AED = 90 °= ∠ ACB, and ∵ ad bisecting ∵ BAC, ∵ DAE = ∠ DAC, ∵ ad = ad, ≌ ACD, ∵ AE = AC, ∵ ad bisecting ≁ BAC, ≁ ad ⊥ CE, that is, the straight line ad is the vertical bisector of the line CE
As shown in the figure, in the triangle ABC, D is a point on the edge AB, and ad = AC, De is parallel to BC, CD bisecting angle EDF is it possible to prove AF bisecting CD bisecting EDF
[find AF vertical bisection CD]
prove:
∵ CD bisection ∠ EDF
∴∠EDC=∠FDC
∵DE//BC
∴∠EDC =∠DCF
∴∠FDC =∠DCF
∴DF =CF
And ∵ ad = AC, AF = AF
∴⊿ADF≌⊿ACF（SSS）
∴∠DAF=∠CAF
The ADC is an isosceles triangle, AF bisects the DAC, according to three lines in one
The AF splits the CD vertically
As shown in the figure, in the triangle ABC, AB > AC, AF is the angular bisector, D is the one on AB, and ad = AC, De is parallel to BC, AC intersection point E, to prove the CD bisector angle EDF
If AF intersects CD with G ∵ ad = AC, and AF is the angular bisector of ∵ CDA ⊥ CD, DG = CG, DF = CF ∵ CDF = DCF and ∵ de parallel BC ∵ CDE = ∵ DCF = ∵ CDF, that is, CD bisector EDF
As shown in the figure, in △ ABC, AB > AC, point D is on AB, ad = AC, de / / BC, CD bisecting ∠ EDF
Picture everybody goes to my space to have a look, in the photo album
∠EDC=∠CDF
De is parallel to BC = > EDC = ∠ DCF
So: ∠ DCF = ∠ CDF = > DF = CF
And because ad = AC, the common edge is AF
So: △ ADF is equal to △ ACF
=> ∠DAF=∠CAF
AF is the height on the bottom edge of the isosceles triangle ADC, so AF bisects the CD vertically
RELATED INFORMATIONS
- 1. As shown in the figure, in the triangle ABC, D is a point on the edge AB, and ad = AC, De is parallel to BC, CD bisection angle EDF proves that AF bisection CD
- 2. In triangle ABC, ad = 3bd, be = 2ec, AF = CF
- 3. As shown in the figure, △ ABC is an equilateral triangle, which is cut by a rectangle parallel to BC, and ab is cut into three equal parts. Then the area of the shadow part in the figure is () A. 19B. 29C. 13D. 49
- 4. (1) if EF = 5cm, then AB = --- (2) what's the difference between AF and de Please help··
- 5. In △ ABC, the angular bisectors AE and be of ∠ BAC and ∠ ABC intersect at point E, extend the circumscribed circle of AE intersecting △ ABC at point D, connect BD, CD and CE, and ∠ BDA = 60 ° ① prove that △ BDE is an equilateral triangle; ② guess what kind of quadrilateral bdce is if ∠ BDC = 120 ° and prove your guess; ③ under the condition of ②, when CE = 4, calculate the area of quadrilateral abdc
- 6. As shown in the figure, point I is the heart of △ ABC, the extension line of AI intersects with BC at point D, and intersects with the circumscribed circle of △ ABC at point E. verify EB = EC = EI
- 7. As shown in the figure, point I is the inner part of triangle ABC, AI intersects BC at point E, and intersection triangle ABC circumscribes circle at point D. prove (1dB = di (2id square = de * ad)
- 8. As shown in the figure, in △ ABC, ∠ C = 90 °, AC = BC, Da bisects ∠ cab and intersects BC at point D. can we determine a point E on AB so that the perimeter of △ BDE equals the length of AB? If yes, please point E and give proof; if not, please explain the reason
- 9. As shown in the figure, ad is perpendicular to D, eg is perpendicular to g, angle e is equal to angle 3, and angle 1 is equal to angle 2
- 10. In the triangle ABC, the angle BAC = 100 degrees, de and FG bisect AB and AC vertically and intersect BC at e and G, respectively
- 11. As shown in the figure, in the triangle ABC, BD bisector angles B and AE are perpendicular to BD and parallel to e and EF, and intersect AB with F
- 12. As shown in the figure, in △ ABC, ab = AC, ad ⊥ BC, ad = AE, if ∠ bad = α, then ∠ EBC is equal to
- 13. In the triangle ABC on the right, DC = 2bd, CE = 3aE, the area of the shadow is 20 square centimeters, and the area of the triangle ABC is______ Cm
- 14. An isosceles triangle is an axisymmetric figure. Its axis of symmetry is the middle line on the bottom edge of a and the high line on the bottom edge of B An isosceles triangle is an axisymmetric figure, and its axis of symmetry is A the center line on the bottom edge B the height on the bottom edge The line on which the bisector of C vertex is located D the line where the height of the waist is
- 15. The middle angle B of triangle ABC is 2 times angle c and ah is perpendicular to BC and h, BM = cm, ab = 2hm
- 16. Let p be a point in Δ ABC, and vector AP = (x-2y) vector AB + (Y-1) vector AC, then the value range of X is--- The value range of Y is-----
- 17. It is known that in △ ABC, ab = AC, D is the midpoint of BC side, P is any point on ad, PE ⊥ AB is in E, PF ⊥ AC is in F. try to explain: (1) PE = pf; (2) Pb = PC
- 18. In △ ABC, ab = AC, ad is the middle line, P is the upper point of AD, CF ∥ AB is made through point C, and the extension BP intersects AC at point E.1, which proves that BP = PC 2, and the square of BP = PE × pf?
- 19. In △ ABC, ad is an angular bisector, BP ⊥ ad is at point P, and ab = 5, BP = 2, AC = 9. It is proved that: ∠ ABC = 3 ∠ C
- 20. Let P = (a + C, b) and q = (B − a, C − a). If P ‖ Q, then the size of angle c is______ ．