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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?
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This is the exterior angle theorem of a triangle. In △ ABH, the sum of inner angle and ∠ AHB + ∠ bad + ∠ Abe = 180 ° and ? bhe is a flat angle, ? AHB + ∠ bad + ∠ Abe = ∠ AHB + ∠ ahe = 180 °, i.e., ∠ bad + ∠ Abe = ∠ ahe
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
Make DM = De, cross AC to m, DN ⊥ AC to point n,
∵DE=DG，
∴DM=DG，
∵ ad is the angular bisector of ∵ ABC, DF ⊥ ab,
∴DF=DN，
In RT △ def and RT △ DMN,
DN=DF
DM=DE
,
∴Rt△DEF≌Rt△DMN（HL），
The area of ∵ △ ADG and △ AED are 50 and 39, respectively,
∴S△MDG=S△ADG-S△ADM=50-39=11，
S△DNM=S△EDF=
One
Two
S△MDG=
One
Two
×11=5.5．
Therefore, B
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,
Ad is the angular bisector of △ ABC, DF ⊥ AB, the perpendicular foot is f, de = DG, the area of △ ADG and △ AED are 80 and 60 respectively, then the area of △ DEF is
D is the angular bisector of △ ABC, DF ⊥ AB, ∵ DF ≔ AB, ∵ DF = DN, def ≡ DNM (HL), the area of adgand △ aedare 80 and 60 respectively,
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
Make DM = De, cross AC to m, DN ⊥ AC to point n,
∵DE=DG，
∴DM=DG，
∵ ad is the angular bisector of ∵ ABC, DF ⊥ ab,
∴DF=DN，
In RT △ def and RT △ DMN,
DN=DF
DM=DE
,
∴Rt△DEF≌Rt△DMN（HL），
The area of ∵ △ ADG and △ AED are 50 and 39, respectively,
∴S△MDG=S△ADG-S△ADM=50-39=11，
S△DNM=S△EDF=
One
Two
S△MDG=
One
Two
×11=5.5．
Therefore, B
As shown in the figure, D is a point on edge ab of △ ABC, DF intersects AC at point E, de = Fe, FC ∥ ab. it is proved that △ ade ≌ △ CFE
Proof: ∵ FC ∥ AB,
∴∠ADE=∠CFE．
In △ ade and △ CFE,
∠ADE=∠CFE，DE=FE，∠AED=∠CEF．
∴△ADE≌△CFE．
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
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∵AE⊥BC，∠EAD=16°，
∴∠ADE=90°-16°=74°．
∵ ade is the external angle of abd, ∵ B = 40 °,
∴∠BAD=∠ADE-∠B=74°-40°=34°．
∵ ad bisection ᙽ BAC to obtain ∠ BAC,
∴∠BAC=2∠BAD=2×34°=68°，
∴∠C=180°-∠BAC-∠B=180°-68°-40°=72°．
Therefore, B
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