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If we make a point F on AB, so that AF=AE, we can know from the bisector of angle BAE that triangle ADE is all equal to triangle ADF, from the midpoint of CD that DF=DE, angle DEA= angle DFA, CE=DE=DF, angle DFB= angle AEC (the complementary angles of equal angles are also equal), and then from angle CAE= angle B that triangle CAE is all equal to triangle DBF.
2.?
Q: As shown in the figure, in the triangle ABC, the bisectors AD, BE and CF intersect at point H, the intersection point H is HG perpendicular to AC, and the vertical foot is G, so is the triangle AHE equal to the triangle CHG? Why? ?
∵∠AHE=∠ABH+∠BAH, (one outer angle of a triangle is equal to the sum of two non-adjacent inner angles);
∠ABH= 1/2∠ABC,
Bah = 1/2 ∠ BAC (meaning angular bisector),
∴∠AHE= 1/2(∠BAC+∠ABC) (equivalent substitution).
∠∠BAC+∠ABC = 180-∠C (the sum of the interior angles of the triangle is equal to 180).
∴∠ahe=( 180-∠ACB)/2 = 90- 1/2∠ACB,
At Rt△HGC,
∠GHC=90 -∠HCG=90 - 1/2∠ACB,
∴? ∠AHE=∠CHG。
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3.? It is known that when △AB=AC, ∠ A = 90, AB=AC, AE⊥BD is in E, ∠ ADB = ∠ CDF, AE is extended from BC to F, which proves that D is the midpoint of AC.
Let d be the symmetry point g of BC connecting FG and CG.
Because angle ADB= angle BAF and angle FDC= angle BAF.
Angle B= angle c = 45.
So angle AFB = 180- angle B- angle BAF = 180- angle C- angle CDF= angle DFG.
So angle AFD+ angle DFG= angle AFD+ angle DFC+ angle AFB = 180.
So lines a, f and G***
And because angle CAG= angle ABD.
Angle ACG = 2 * 45 = 90 = bad angle.
So all the bad triangles are equal to triangle ACG.
So CG=AD
CG=DC
So AD=DC
4.? In the known triangle ABC, AD is the center line of BC side, E is the upper point of AC, BE and AD intersect at F, and if AE=EF, it is verified that AC=BF.
Expand AD to m so that DM = ad, even BM, CM.
AD=DM,BD=CD
∴ABMC is a parallelogram (diagonal bisection)
∴AC‖BM,AC=BM (equal to the last use)
∴∠DAC=∠DMB(∠DAC is ∠EAF, ∠DMB is ∝∞∠bmf used below) (the internal angles are equal) ...
In the triangle AEF,
AE = EF
∴∠EAF=∠EFA (isosceles triangle) ... ②
∫∠EFA =∠BFM (relative vertex angle is equal) ... ③
From ① ② ③, we get ∠EAF=∠EFA=∠BFM=∠BMF.
In a triangular BFM,
∠∠BFM =∠BMF
∴ Triangle BF=BM is an isosceles triangle with side length BF = BM.
AC=BM,AC = BF。
5.? Given triangle ABC, AD is the center line on the side of BC, E is a point on AC, AD and BE intersect at point F, AE=EF. BF=AC?
Parallel lines that extend AD and pass through point b and are AC intersect at point g.
Then AC//BG, AE=EF,
Available BF=BG
In triangle BDG and triangle CDA
BD=CD,& ltADC = & ltGDB,& ltDBG = & lt; ACD,
Consistency of two triangles
So AC=BG=BF