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As shown in the figure, it is known that ad is the bisector angle of the isosceles triangle ABC, and the angle c = 90 degrees proves that ab = AC + CD
It is proved that through point D, do de, perpendicular AB, and cross AB to E
The distance between the point on the bisector and the two sides of the corner is equal: CD = De
And the triangle ACD is congruent with the triangle ade, then AC = AE
Because the triangle ABC is an isosceles right triangle, the angle B = 45 ° and because De is perpendicular to AB, we can get de = EB
So AB = AE + EB = AC + de = AC + CD
Get the certificate
As shown in the figure, in △ ABC, the angular bisectors ad, be and CF intersect at point h, passing through point h as Hg ⊥ AC, and the foot perpendicular to g, then ∠ ahe = ∠ CHG? Why?
Reason: ∵ ad, be and CF are the angular bisectors of ad, be, CF are the angular bisectors of △ ABC, bad = ∠ CAD = x, ∠ Abe = ∠ CBE = y, ∠ BCF = ∠ ACF = Z, then 2x + 2Y + 2Z = 180 ° i.e. x + y + Z = 90 ° in △ AHB, ? ahe is the outer angle of △ AHB, ? ahe = ∠ bad + ∠ Abe = x + y = 90 ° - Z
As shown in the figure, in △ ABC, the angular bisectors ad, be and CF intersect at point h, passing through point h as Hg ⊥ AC, and the foot perpendicular to g, then ∠ ahe = ∠ CHG? Why?
∠AHE=∠CHG．
Reason: ∵ ad, be, CF are the angular bisectors of ᙽ ABC,
ν can be set ∠ bad = ∠ CAD = x, ∠ Abe = ∠ CBE = y, ∠ BCF = ∠ ACF = Z,
Then 2x + 2Y + 2Z = 180 °,
That is, x + y + Z = 90 °,
In △ AHB,
∵ ahe is the outer angle of AHB,
∴∠AHE=∠BAD+∠ABE=x+y=90°-z，
In △ CHG, ∠ CHG = 90 ° - Z,
∴∠AHE=∠CHG．
As shown in the figure, in △ ABC, the angular bisectors ad, be and CF intersect at point h, passing through point h as Hg ⊥ AC, and the foot perpendicular to g, then ∠ ahe = ∠ CHG? Why?
Reason: ∵ ad, be and CF are the angular bisectors of ad, be, CF are the angular bisectors of △ ABC, bad = ∠ CAD = x, ∠ Abe = ∠ CBE = y, ∠ BCF = ∠ ACF = Z, then 2x + 2Y + 2Z = 180 ° i.e. x + y + Z = 90 ° in △ AHB, ? ahe is the outer angle of △ AHB, ? ahe = ∠ bad + ∠ Abe = x + y = 90 ° - Z
As shown in Figure 5, in triangle ABC, the bisectors ad, be and CF intersect at point h, Hg is perpendicular to AC and g. imagine the relationship between angle ahe and angle CHG, and prove your conjecture Grade emergency
Because ad, be, CF are angular bisectors
therefore
∠BAD＝∠BAC/2
∠ABE＝∠ABC/2
∠ACF＝∠ACB/2
therefore
∠AHE＝∠BAD＋∠ABE
＝∠BAC/2＋∠ABC/2
＝(∠BAC＋∠ABC)/2
＝(180°－∠BCA)/2
＝90°－∠BCA/2
＝90°－∠ACF
＝90°－∠GCH
Because he ⊥ AC
So ∠ CHG = 90 ° - ∠ GCH
Therefore, ahe = CHG
As shown in the figure, in △ ABC, the angular bisectors ad, be and CF intersect at point h, passing through point h as Hg ⊥ AC, and the foot perpendicular to g, then ∠ ahe = ∠ CHG? Why?
Reason: ∵ ad, be and CF are the angular bisectors of ad, be, CF are the angular bisectors of △ ABC, bad = ∠ CAD = x, ∠ Abe = ∠ CBE = y, ∠ BCF = ∠ ACF = Z, then 2x + 2Y + 2Z = 180 ° i.e. x + y + Z = 90 ° in △ AHB, ? ahe is the outer angle of △ AHB, ? ahe = ∠ bad + ∠ Abe = x + y = 90 ° - Z
As shown in the figure, ad is the angular bisector of △ ABC, DF ⊥ AB, the perpendicular foot is f, de = DG, the areas of △ ADG and △ AED are 50 and 39 respectively, then the area of △ EDF is () A. 11 B. 5.5 C. 7 D. 3.5
Let DM = De, cross AC to m, DN ⊥ AC at point n, ∵ de = DG, ∵ DM = DG, ∵ ad is the angular bisector of ⊥ ABC, DF ⊥ AB, ᙽ DF = DN. In RT △ def and RT △ DMN, DN = DF = DM = De, ≌ RT ≌ DMN (HL), ∵ ADG and △ AED are 50 and 39 respectively,
As shown in the figure, D is a point AB / / FC on the edge ab of the triangle ABC, and DF intersects AC at point e. it is proved that AE is equal to CE
It seems that the conditions are not enough
Only the upper and lower triangles can be proved to be similar
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
Proof: as shown in the figure, extend Fe to g, make eg = EF, connect CG
In △ def and △ CEG,
A kind of
ED＝EC
∠DEF＝∠CEG
FE＝EG ，
∴△DEF≌△CEG．
∴DF=GC，∠DFE=∠G．
∵DF∥AB，
∴∠DFE=∠BAE．
∵DF=AC，
∴GC=AC．
∴∠G=∠CAE．
∴∠BAE=∠CAE．
That is, AE bisection ∠ BAC
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
Proof: as shown in the figure, extend Fe to g, make eg = EF, connect CG
In △ def and △ CEG,
A kind of
ED＝EC
∠DEF＝∠CEG
FE＝EG ，
∴△DEF≌△CEG．
∴DF=GC，∠DFE=∠G．
∵DF∥AB，
∴∠DFE=∠BAE．
∵DF=AC，
∴GC=AC．
∴∠G=∠CAE．
∴∠BAE=∠CAE．
That is, AE bisection ∠ BAC
RELATED INFORMATIONS
- 1. It is known that, as shown in the figure, in △ ABC, ad and AE are the bisectors of the height and angle of △ ABC respectively. If ∠ B = 30 ° and ∠ C = 50 °, calculate the degree of ∠ DAE
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- 8. As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = AC, points D and E are on BC and AC respectively, and BD = CE, M is the midpoint of AB, and △ MDE is isosceles triangle As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = AC, points D and E are respectively on BC and AC, and BD = CE, M is the midpoint of AB, and △ MDE is isosceles right triangle. Please explain the reasons
- 9. As shown in the figure, in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at D, e is a point on AB, AF ⊥ CE is at F, ad intersects CE at G point. It is proved that ∠ B = ∠ CFD
- 10. As shown in the figure, △ ABC, ab = AC, points D and E are on the extension line of AB and AC respectively, and BD = CE, de and BC intersect at point F. verification: DF = EF
- 11. Ad is the height of the triangle ABC, and the circle O intersects with AD as its diameter AB.AC At point E. F, it is proved that angle B is equal to angle AFE
- 12. As shown in the figure, in △ ABC, the angle BAC = 90 degrees, ad ⊥ BC bisects ∠ ABC at point D and BF, then △ AEF is an isosceles triangle? And the proof is given
- 13. In the equilateral triangle ABC, ad and CE are the heights on BC and ab respectively, and AD and CE intersect at point F, what is the degree of ∠ AFE? If points D and e run on BC and ab respectively, in order to make the conclusion (1) hold, please guess what quantitative relationship BD and AE should satisfy and give proof
- 14. It is proved that ab minus AC is greater than BD minus DC
- 15. As shown in the figure, in the triangle ABC, ad is perpendicular to BC, CE is perpendicular to AB, BC = 6cm, ab = 8cm, calculate the value of AD: CE
- 16. As shown in the figure, in triangle ABC, angle B is equal to angle c, points D, e, f are on AB, BC and AC respectively, and BD = CE, angle def = angle B, whether there is and triangle BD in the graph
- 17. As shown in the figure, B is a point on the line ad, △ ABC and △ BDE are equilateral triangles, which connect CE and extend the extension line crossing ad at point F, and the circumscribed circle ⊙ o of ⊙ ABC intersects CF at point P (1) It is proved that be is the tangent of ⊙ o; (2) If CP = 2, PF = 8, find the length of AC
- 18. Let D, e, f be the midpoint of BC, AC and ab of the triangle ABC respectively How to do such a topic,
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