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As shown in the figure, it is known that AE is the center line of triangle ABC, and the perimeter of triangle Abe is 2cm more than that of triangle ace, if AB + A If AB + AC = 18cm, request AB, AC length
If the straight line L is parallel to AB and CD through point E, then ∠ bed is divided into two legs, the upper one is equal to ∠ Abe, and the lower one is always equal to ∠ CDE, because BF bisection ∠ Abe, DF bisection ∠ CDE, we can get ∠ FBE + ∠ EDF = 75 ° / 2 = 37.5 ° according to the sum of inner angles of quadrilateral is equal to 360 °, we can know that ∠ BFD = 37.5 °
As shown in the figure, in △ ABC, ∠ cab = 90 °, ad ⊥ BC in D, be is the bisector of ∠ ABC, crossing ad at F, and verification: DF / AF = AE / EC
It is proved that: to do eg ? BC in G ≓ be is the bisector of the angle B ᚛ AE = Ge. (1) ? ad ? BC ? ad ? BC 8857; ad ? BC ? B ⊙ angle Fab = 90 ° - angle B = angle c = angle CBE = angle Abf ⊙ CBE ⊙ AF / EC = BF / be. (2) ? BC, eg 8 BC ? BDF ? Bge X DF / Ge ⊙ Ge ⊙ f it's a good idea
As shown in Fig. 3, it is known that in △ ABC, ∠ cab = 90 °, ad ⊥ BC, AE = EC, the extension line of AB is F. it is proved that AB / AC = DF / fa
The triangle ABC is similar to abd (right angle, ∠ cab common), AB / AC = BD / ad ∠ BCA = ∠ bade is the ACD center line of right triangle, de = AC / 2 = CE, ∠ EDC = ∠ BCA = ∠ bad because ∠ BDF = ∠ EDC (opposite top), so ∠ BDF = ∠ bad. ∠ AFD is similar to AFD, and the result is: DF / FA = BD / ad =
As shown in the figure, in △ ABC, the bisectors AD and CE of ∠ B = 60 °, BAC, ∠ ACB intersect at point O, which shows the reason why AE + CD = AC
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As shown in the figure, in △ ABC, the bisectors AD and CE of ∠ B = 60 °, BAC, ∠ ACB intersect at point O, which shows the reason why AE + CD = AC
It is proved that if AF = AE is taken on AC and connected with of, then △ AEO ≌ △ AFO (SAS), ∵ AOE = ∠ AOF; ∵ AD and CE are bisected ∠ BAC, ∠ ACB,
In △ ABC, the angle bisector AD and CE of ∠ B = 60 ° a C intersect with point O, and the relationship between line segments AE, CD and AC is conjectured and proved I want to write the process concretely. If I write it well, I will give you 50 points and 100 points. I hope I can write it concretely
AC=AE+CD
You take a little h on AC to make AE = ah
Prove again △ AEO ≌ △ aho
So AE = ah
Prove again △ CHO ≌ △ CDO
Then CD = ch
So AC = ah + CH = AE + CD
In this paper, the area of a right triangle in ABC is 2 cm?
If s △ ADC = 12 · ab · DC = 12 × 5 × 2 = 5 (square centimeter), then s △ ABF = x, s △ AFE = y, then s △ ADF = (s △ ADC △ 2) - y = 2.5-y, so bFEF = xy = 42.5 − y （1）X+Y=4+5-Y⇒X=9-2Y… （2）...
As shown in the figure, in the triangle ABC, BD = DC, AE = EF, verification: BF = AC
prove:
Method 1: extend ad to point m, make MD = FD, connect MC,
∴△BDF≌CDM（SAS）．
∴MC=BF,∠M=∠BFM．
∵EA=EF,
∴∠EAF=∠EFA,
∵∠AFE=∠BFM,
∴∠M=∠MAC,
∴AC=MC,
∴BF=AC
Method 2: extend ad to m, make DM = ad, connect BM,
∴△ADC≌△MDB（SAS）,
∴∠M=∠MAC,BM=AC,
∵EA=EF,
Ψ cam = AFE, and ∠ AFE = ∠ BFM,
∴∠M=∠BFM,
∴BM=BF,
∴BF=AC．
Hope to adopt, thank you
It is known that in the triangle ABC, ab = AC, extending Ba to D, AE bisecting ∠ DAC, AE parallel to BC, why?
Because D is the upper point of the Ba extension,
Then ∠ DAC = ∠ B + ∠ C
Because AB = AC
So ∠ B = ∠ C
So ∠ DAC = 2 ∠ C
Because AE bisects ∠ DAC
Then ∠ EAC = 1 / 2 ∠ DAC = ∠ C
So AE ‖ BC
As shown in the figure, ad = dB, AE = EF = FC, the area of triangle DEF is 5 square centimeters, and what is the area of triangle ABC?
The triangle ade is 1 / 3 of the triangle Abe area (because AB = 3aD), so the triangle ade is (1 / 3) * (1 / 3) = 1 / 9 54 * (1 / 9) = 6 of triangle ABC, so it is 6 square centimeter primary problem
RELATED INFORMATIONS
- 1. As shown in the figure AB = ad ∠ bad = ∠ CAE AC = AE, ab = ad
- 2. As shown in the figure, triangle ABC is all equal to triangle ade, if angle BAE = 120 ° angle bad = 40 ° angle BAC =?
- 3. In △ ABC, a = 2, B = 2. If the triangle has a solution, what is the value range of a? A belongs to (0 degree, 90 degree). Why not?
- 4. As shown in the figure, points D and E are on AB and AC respectively, and ∠ ABC = ∠ AED. If de = 4, AE = 5, BC = 8, then the length of AB is______ .
- 5. In trapezoid ABCD, AD / / BC, ad = 3, BC = 5, points E and F are on AB and DC respectively, EF / / BC, if AE: EB = 2:3, find EF I also calculated 3. But the answer is 2
- 6. In △ ABC, D is a point on the edge of BC, BD = 1 2dc, ∠ ADB = 120 ° ad = 2, if the area of △ ADC is 3, then AB = () A. 1 B. Five C. Seven D. 2 Two
- 7. Given the triangle ABC, if ∠ C = 90 °, the square of 4B + the square of 3C = four times the root sign 3bC, find the value of cosa
- 8. In the right triangle ABC, ∠ C = 90 °, BC: AC = 1: radical 3, CD ⊥ AB in D, find s triangle CDB: s triangle ABC
- 9. In the RT triangle ABC, the angle ACB = 90 degrees, ab = 4, take AC and BC as the diameters to make semicircles, and the areas are denoted as S1 and S2 respectively?
- 10. In the RT triangle ABC, ∠ C = 90 °, ab = radical 10, AC: BC = 2:1, calculate the perimeter and area of RT triangle
- 11. The known points D and E are on the sides AB and AC of the triangle ABC, ab = 10cm, AC = 6cm, ad = 2cm=__ Triangle ade is similar to triangle ABC Since there is no drawing, there will be two situations I have one more question to answer
- 12. In the triangle ABC, D is a point on the edge of AC, the angle DBC = angle a, BC = radical 6, AC = 3, find the length of CD Urgent``````
- 13. As shown in the figure, in the RT triangle ABC, the angle c = 90 degrees, the angle ABC = 30 degrees, the bisecting angle BAC BD, and the parallel AC prove that AE = be As shown in the figure, the angle c = 90 degree angle ABC = 30 degree ad bisection angle BAC BD parallel AC (1) proves AE = be (2) Verify BC + CE = de
- 14. As shown in the figure, in triangle ABC, angle ACB = 90 degrees, AE = AC, BC = BD, calculate the degree of angle ECD Ed on ab
- 15. As shown in the figure, in △ ABC, ad is an angular bisector, e is a point on ad, and CE = CD. If AE = 2DE, the area of △ ace is 6, calculate the area of △ ABC
- 16. In RT △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of BC, e and F are on AC and AB, and AE = BF. It is proved that △ DEF is an isosceles right triangle
- 17. It is known that: as shown in the figure, △ ABC (AB ≠ AC), D and E are on BC, and de = EC, DF ‖ Ba is made through D, and AE is intersected with point F, DF = AC. it is proved that AE bisection ∠ BAC
- 18. As shown in the figure, △ ABC, D is on AC, ad: DC = 1:2, e is the midpoint of BD, and the extension line of AE intersects BC at F, Verification: BF: FC = 1:3
- 19. In the triangle ABC, if a + C = 2B, C = 2, a = radical 3 + 1, find the value of B as in the title
- 20. In the triangle ABC, AB is equal to AC, angle a is equal to 20 degrees, and point D is on AB, so that ad is equal to BC, and the degree of angle BDC is obtained Note: this problem only isosceles triangle and congruent triangle to prove