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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
DC: 1: proof,
∴AD：AC=1：3．
Make DG parallel to AF and BC to g, then CD
CA=GC
CF，
According to the nature of proportion, ad
AC=FG
FC=1
3，
And E is the midpoint of BD,
The EF is the median line of △ bgd,
∴BF=FG．
∴BF
FC=1
BF: FC = 1:3
In the triangle ABC, D is on AC, ad: DC = 1:2, e is the midpoint of BD, the extension line of AE intersects BC at F, and BF: FC = 1:3 It's better to use the principle of similar triangles
It is proved that: take the midpoint m of DC, make the parallel lines of AF through D and m, and cross BC and G and N respectively
From the fact that D and m are the trisection points of AC, FG = GN = NC
If e is the midpoint of BD, BF = FG
So: BF = FG = GN = NC
So: BF: FC = 1:3
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
DC: 1: proof,
∴AD：AC=1：3．
Make DG parallel to AF and BC to g, then CD
CA=GC
CF，
According to the nature of proportion, ad
AC=FG
FC=1
3，
And E is the midpoint of BD,
The EF is the median line of △ bgd,
∴BF=FG．
∴BF
FC=1
BF: FC = 1:3
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
DC: 1: proof,
∴AD：AC=1：3．
Make DG parallel to AF and BC to g, then CD
CA=GC
CF，
According to the nature of proportion, ad
AC=FG
FC=1
3，
And E is the midpoint of BD,
The EF is the median line of △ bgd,
∴BF=FG．
∴BF
FC=1
BF: FC = 1:3
As shown in the figure, D is the midpoint of BC of triangle ABC, f is the midpoint of AD, and the extension line of BF intersects AC at point E, thus AE = 1 / 2ce
It is proved that DM / / be is cross AC to e BD = CD cm = EM EF / / DM AF = DF AE = EM
AE=EM=MC AE=1/2CE
As shown in the figure, △ ABC is an equilateral triangle, D and E are two points on AC and BC respectively, ad = CE, and AE and BD intersect at point P, BF ⊥ AE at point F. if BP = 6, find the length of PF
∵△ ABC is an equilateral triangle, AB = AC, ∠ BAC = ∠ C, in △ abd and △ CAE, ab = AC ∠ bad = ∠ CAD = CE ≌△ CAE (SAS), abd = ∠ CAE, APD = ∠ ABP + ∠ PAB = ∠ BAC = 60 °, and ≌ BPF = ∠ APD = 60 ° in RT △ BFP, ∠ PBF = 30 °, BP = 2pF
In △ ABC, f is the midpoint of AC. if BC and BF of D and E are crossed with P and Q respectively with AD and AE, then BP: PQ: QF = () A. 5：3：2 B. 3：2：1 C. 4：3：1 D. 4：3：2
F is used to make FN ‖ BC, AE to m, ad to n,
∵ f is the midpoint of AC,
/ / FM is the median line of △ AEC,
∴MF=1
2CE，CE=2FM，
∵BD=DE=CE，
∴BE=2CE=4FM，
∵FM∥BC，
∴△FMQ∽△BEQ，
∴FQ
BQ=FM
BE=1
4，
∵ FN is the median line of ∵ ADC,
∴FN=1
2CD=CE=BD，
∵FN∥BC，
∴△FNP∽△BDP，
∴BP
PF=BD
FN=1，
∴BP=PF，
∵FQ
BQ=1
4，
∴FQ
BF=1
5，
∴FQ=1
5BF，
∵BP=1
2BF，FQ=1
5BF，
∴PQ=PF-QF=1
2BF-1
5BF=3
10BF，
∴BP：PQ：QF=（1
2BF）：（3
10BF）：（1
5BF）=5：3：2．
Therefore, a
In the triangle ABC, D is the midpoint of BC, DF intersects AC at point E, and the extension line of Ba is at point F. it is proved that AE: CE = AF: BF
Make a parallel line of AC through point B, and the extension line of crossing FD is h, CD = BD, so BH = CE, AE: BH = AF: BF can be seen at a glance
As shown in the figure, in △ ABC, ab = AC, D is the midpoint of BC, DF ⊥ AC, e is the midpoint of DF, and AE and BF are connected
It is proved that: (1) take the midpoint g of CF to connect DG and da,
∵ D is the midpoint of BC, ab = AC,
∴AD⊥BC
,
∵DF⊥AC，
∴∠DAF=∠FDC，
∴△DAF∽△DFC，
∴AF：DF=DF：CF，
∴DF2=CF•AF；
(2) ∵ e is the midpoint of DF and G is the midpoint of FC,
∴AF：DF=EF：FG，
∴△AFE∽△DFG，
∴∠FAE=∠FDG，
∵ G is the midpoint of FC
In △ CBF, DG ∥ BF,
∴∠GDF=∠BFD，
∴∠FAE=∠BFD，
∵AF⊥DF，
∴∠FAE+∠FEA=90°，
∴∠BFD+∠FEA=90°，
∴AE⊥BF．
It is known that in the triangle ABC, point D is the midpoint of the AB side, AE ⊥ BC, BF ⊥ AC, and the perpendicular feet are points e, F, AE, BF, which intersect with point m and connect de and DF. If de = k × DF, what is the value of K?
K = 1,2 floor did not understand, why it must be a right triangle? Acute angle triangle also meets this condition
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