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As shown in the figure, ab = AC, ad = AE are known
It is proved that AF ⊥ BC is applied to F,
∵ AB = AC (known),
ν BF = CF (three wires in one),
And ∵ ad = AE (known),
/ / DF = ef (three wires in one),
ν bf-df = cf-ef, i.e. BD = Ce (property of the equation)
Known: in the triangle ABC, ab = AC, D is the point on BC, the angle bad = 50 degrees, e is the point on AC, AE = ad, calculate the degree of angle EDC
According to the conditions, it can be concluded that:
＜ABC=＜ACB, ＜ADE=＜AED
As shown in the figure: < AED = < ACB + < EDC
＜EDC=＜AED-＜ACB
＜DAC+50+＜ABC+＜ACB=180
＜DAC+50+2＜ACB=180
＜DAC=130-2＜ACB
Also: < DAC + < ade + < AED = 180
＜DAC+ 2＜AED=180
130-2＜ACB+ 2＜ADE=180
2(＜AED-＜ACB）＝180-130=50
＜EDC=50/2=25
The second solution is as follows
＜ADE+＜EDC=＜BAD+＜ABC
=＜BAD+＜ACB
＜AED=＜EDC+＜ACB
＜ADE=＜EDC+＜ACB
＜ADE-＜EDC=＜ACB
＜ADE+＜EDC=＜BAD+＜ACB
2＜EDC=＜BAD
＜EDC=50
The third solution: hypothesis method
Suppose DAC = 60,
Then < ade = 60
＜ABC=（180-50-60）/2=35
＜EDC+＜ADE=＜ABC+50
＜EDC=＜ABC+50-＜ADE
=35-50-60
=25
As shown in the figure, in the parallelogram ABCD, e is a point on the edge of BC, and ab = AE (1) Verification: △ ABC ≌ △ ead; (2) If AE bisects ∠ DAB, ∠ EAC = 25 °, find the degree of ∠ AED
(1) It is proved that: ∵ quadrilateral ABCD is a parallelogram,
∴AD∥BC，AD=BC．
∴∠DAE=∠AEB．
∵AB=AE，
∴∠AEB=∠B．
∴∠B=∠DAE．
∵ in △ ABC and △ AED,
AB＝AE
∠B＝∠DAE
AD＝BC ，
∴△ABC≌△EAD．
(2) ∵ AE bisection ∵ DAB (known),
∴∠DAE=∠BAE；
And ∵ DAE = ∵ AEB,
∴∠BAE=∠AEB=∠B．
The △ Abe is an equilateral triangle
∴∠BAE=60°．
∵∠EAC=25°，
∴∠BAC=85°．
∵△ABC≌△EAD，
∴∠AED=∠BAC=85°．
As shown in the figure, in the parallelogram ABCD, e is a point on the edge of BC, and ab = AE (1) Verification: △ ABC ≌ △ ead; (2) If AE bisects ∠ DAB, ∠ EAC = 25 °, find the degree of ∠ AED
(1) It is proved that: ∵ quadrilateral ABCD is a parallelogram,
∴AD∥BC，AD=BC．
∴∠DAE=∠AEB．
∵AB=AE，
∴∠AEB=∠B．
∴∠B=∠DAE．
∵ in △ ABC and △ AED,
AB＝AE
∠B＝∠DAE
AD＝BC ，
∴△ABC≌△EAD．
(2) ∵ AE bisection ∵ DAB (known),
∴∠DAE=∠BAE；
And ∵ DAE = ∵ AEB,
∴∠BAE=∠AEB=∠B．
The △ Abe is an equilateral triangle
∴∠BAE=60°．
∵∠EAC=25°，
∴∠BAC=85°．
∵△ABC≌△EAD，
∴∠AED=∠BAC=85°．
As shown in the figure, in the parallelogram ABCD, e is a point on the edge of BC, and ab = AE (1) Verification: △ ABC ≌ △ ead; (2) If AE bisects ∠ DAB, ∠ EAC = 25 °, find the degree of ∠ AED
(1) It is proved that:
In the triangle ABC, D is the midpoint of BC, De is perpendicular to the intersection angle of BC, AE of BAC is perpendicular to e, EF is perpendicular to AB and F, eg is perpendicular to AC, and the extension of AC is perpendicular to g It is proved that BF = CG
In RT △ AFE and △ age,
Angle FAE = angle EAG
AE = AE (public)
All △ AFE is equal to △ age (ASA) (because in RT △ a slanted edge is equal to an acute angle, and the other acute angle must be equal)
Therefore, EF = eg (the corresponding sides of an congruent triangle are equal)
Because De is perpendicular to point D and D is the midpoint of BC, Δ BEC is an isosceles triangle, be = CE
In RT △ BFE and △ CGE, EF = eg, be = CE
Then, RT △ BFE is all equal to RT △ CGE (ASA, or SAS) (in RT △ there are two corresponding equal sides, and their corresponding acute angles must be equal)
Therefore, BF = CG (the corresponding sides of an congruent triangle are equal)
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
Proof: connect be, EC,
∵ED⊥BC，
D is the midpoint of BC,
∴BE=EC，
∵EF⊥AB EG⊥AG，
And AE bisection ∠ fag,
∴FE=EG，
In RT △ BFE and RT △ CGE,
BE＝CE
EF＝EG ，
∴Rt△BFE≌Rt△CGE （HL），
∴BF=CG．
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
Proof: connect be, EC,
∵ED⊥BC，
D is the midpoint of BC,
∴BE=EC，
∵EF⊥AB EG⊥AG，
And AE bisection ∠ fag,
∴FE=EG，
In RT △ BFE and RT △ CGE,
BE＝CE
EF＝EG ，
∴Rt△BFE≌Rt△CGE （HL），
∴BF=CG．
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
Proof: connect be, EC,
∵ED⊥BC，
D is the midpoint of BC,
∴BE=EC，
∵EF⊥AB EG⊥AG，
And AE bisection ∠ fag,
∴FE=EG，
In RT △ BFE and RT △ CGE,
BE＝CE
EF＝EG ，
∴Rt△BFE≌Rt△CGE （HL），
∴BF=CG．
As shown in the figure, D and E are the points on the ABC sides BC and AC of the triangle. AD and be intersect with F, and BF = EF, AE = 3ec, Find the value of BD: DC
Eg ∥ BC is made by E and handed over to ad to g,
Then eg / CD = AE / AC = 3 / 4, ν CD = 4 / 3eg,
∵BF=EF,∴EG/BD=EF/BD=1,∴EG=BD,
∴BD：DC=EG：4/3EG=3：4.
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