It is known that in △ ABC and △ def, am and DN are respectively in BC and EF, and AB / de = am / dN = BC / EF. It is proved that △ ABC is similar to △ def

It is known that in △ ABC and △ def, am and DN are respectively in BC and EF, and AB / de = am / dN = BC / EF. It is proved that △ ABC is similar to △ def

It is known that in △ ABC and △ def, am and DN are the midlines on BC and EF respectively, and AB / de = am / dN = BC / EF. It is proved that △ ABC is similar to △ def
It is proved that: ∵ m and N are the midpoint of BC and EF respectively, ∵ BC / EF = 2bm / 2EN = BM / en;
So AB / de = am / dN = BM / en, ∑△ ABM ～△ den, (the three sides are proportional, the two triangles are similar)
Two triangles are similar: one angle is equal, two sides are proportional, and two triangles are similar

It is known that in △ ABC and △ def, ab = De, AC = DF, am and DN are midlines respectively, and am = DN, so it is proved that △ ABC is congruent to △ def

Extend am to P, make MP = am, connect BP,
Extend DN to Q, make QN = DN, connect EQ,
∵BM=CM,∠ANC=∠BMP,
∴ΔAMC≌ΔPMB,
∴AC=BP,∠MAC=∠P,
Similarly, DF = EQ, ∠ NDF = q,
∵AB=DE,AC=DF,AM=DN,
∴BP=EQ,AP=DQ,
∴ΔABP≌ΔDEQ,
∴∠BAM=∠EDN,∠P=∠Q,
∴∠MAC=∠NDF,
∴∠BAM+∠MAC=∠EDN+∠NDF,
That is, BAC = EDF,
And ab = De, AC = DF,
∴ΔABC≌ΔDEF(SAS).

It is known that: in △ ABC and △ def, ab = De, BC = EF, am is the middle line of △ ABC, DN is the middle line of △ def, am = DN, the proof is △ ABC ≌ Δ def

It is proved that: ∵ BC = EF, am is the middle line of △ ABC, DN is the middle line of △ def, ∵ BM = en, in △ ABM and △ den, ab = Adam = dnbm = en ≌ ABM ≌ Δ den, ≌ B = ≌ e, in △ ABC and △ def, ab = de ≌ B = ≌ EBC = EF ≌ ABC ≌ def

It is known that am is the middle line on the edge BC in △ ABC, D and E are the points on AB and AC respectively, and ad = AE, de intersects am in N. verification: DN × AB = en × AC

Extend am to f so that MF = am
Then BF / / AC, and BF = AC
∴ ∠ABF=180°-∠BAC=2∠ADE
Crossing AB with P through n as NP / BF
∴ ∠APN=∠ABF=2∠ADE
∴ ∠PND= ∠PDN
PN：BF=AN：AF……………… ①
Passing through n as NQ / / CF to AC to Q
∴ ∠AQN=∠ACF=2∠AED
∴ ∠QNE= ∠QEN=∠PND= ∠PDN
QN：CF=AN：AF……………… ②
According to ① and ②, PN: BF = QN: CF
∴ PN：QN=BF：CF…………… ③
∵ ∠QNE= ∠QEN=∠PND= ∠PDN
∴ △PDN∽△QEN
∴ PN：NQ=DN：NE…………… ④
It can be concluded from (3) and (4) that BF: CF = DN: ne
Namely: AC: ab = DN: ne
∴ DN×AB=EN×AC
(Note: the area method is quite simple.)

In △ ABC, if am is the middle line on the edge of BC, D is a point on BM, the parallel line passing through D as am intersects AB at point E, and the extension line of Ca intersects at point F, then the equation de + DF = 2am must hold. Please explain the reason

Because am is the middle line on the edge of BC, then BG = De, BP = 2am, go = BM = cm = (1 / 2) BC, ∵ BM = cm, GH ∥ BC ∥ OE = Oh (triangle similarity, OE / BM = AO / am = Oh / cm...)

As shown in the figure, am is the middle line of △ ABC, and de ‖ BC intersects AB and AC to D and e respectively, and intersects am to N. verify: DN = en

The narration is not clear, the topic has a problem, you see for yourself!

Geometry in the triangle ABC, ab = AC, D is a point on the extension line of Ca, De is perpendicular to BC at point E, AB intersects at point F, proving that the triangle ADF is an isosceles triangle

Because De is vertical to BC
So Feb is a right triangle
In the same way
Dec is a right triangle
So angle DFA = angle BFE = 90 ° - angle B
Angle FDA = angle EDC = 90 ° - angle c
Because AB = AC
So angle B = angle C
So angle DFA = angle FDA
Because ADF is three vertices of the same triangle
therefore
When angle DFA = angle FDA
Edge ad = edge AF
So triangle ADF is isosceles triangle

In the triangle ABC, ab = AC, D is the point on AB, through d make de ⊥ BC at e, and intersect with the extension line of Ca at F, judge the shape of triangle ADF and explain
Make the process clear

∵AB=AC
∴∠B＝∠C
⊙ de ⊥ BC in e
∴∠BDE＝90°－∠B
∠CFE＝90°－∠C
∴∠BDE＝∠CFE
∵∠BDE＝∠FDA
∴∠AFD＝∠FDA
∴AF=DF
Triangle ADF isosceles triangle

In △ ABC, ab = AC, D is the point on AB, passing through D is de ⊥ BC, the perpendicular foot is e, and it intersects with the extension line of Ca at point F. it is proved that △ ADF is an isosceles triangle

∵DE⊥BC
∴∠FEC=∠BED=90°
∴∠F+∠C=90°,∠B+∠BDE=90°
∵AB=AC,
∴∠B=∠C
∴∠F=∠BDE
∵∠BDE=∠FDA
∴∠F=∠FDA
∴FA=AD
The ∧ ADF is an isosceles triangle
It's been written for a long time. I promise that's right!

As shown in the figure, it is known that in △ ABC, ab = AC, D is the point on AB, de ⊥ BC, e is the perpendicular foot, the extension line of ED intersects the extension line of Ca at point F, and the proof is: ad = AF

It is proved that: ∵ AB = AC, ∵ B = ∵ C, ∵ de ⊥ BC, ∵ C + F = 90 °, B + BDE = 90 °, ∵ ADF = BDE, ∵ f = ADF, ∵ ad = AF