D is the middle of the edge BC of the triangle ABC, De is vertical AC, DF is vertical AB, the vertical feet are e, F, and BF = CE, judge the shape of triangle ABC Is it an isosceles right triangle? Give proof

D is the middle of the edge BC of the triangle ABC, De is vertical AC, DF is vertical AB, the vertical feet are e, F, and BF = CE, judge the shape of triangle ABC Is it an isosceles right triangle? Give proof

It is known that D is the midpoint of BC
So: BD = CD
Known: BF = CE
So: right triangle BDF is equal to right triangle CDE
So: ∠ FBD = ∠ ECD
So the triangle ABC is an isosceles triangle

As shown in the figure: the bisector BF of ∠ ABC intersects with the bisector CF of the adjacent external angle of ∠ ACB in △ ABC, and passes through F as DF ‖ BC, crosses AB at D, crosses AC at e, extends BC to m, then: ① How many isosceles triangles are there? Why? ② What is the relationship between BD, CE and de? Please prove it

(1) There are two isosceles triangles, namely △ BDF and △ CEF,
∵ BF and CF bisect the external angles of ᙽ ABC and ᙽ ACB respectively,
∴∠DBF=∠CBF，∠FCE=∠FCM，
∵DE∥BC，
∴∠DFB=∠CBF，∠EFC=∠FCM，
∴∠DBF=∠DFB，∠FCE=∠EFC，
∴BD=FD，EF=CE，
△ BDF and △ CEF are isosceles triangles;
(2) Existence: bd-ce = De,
Proof: DF = BD, CE = EF,
∴BD-CE=FD-EF=DE．

As shown in the figure: the bisector BF of ∠ ABC intersects with the bisector CF of the adjacent external angle of ∠ ACB in △ ABC, and passes through F as DF ‖ BC, crosses AB at D, crosses AC at e, extends BC to m, then: ① How many isosceles triangles are there? Why? ② What is the relationship between BD, CE and de? Please prove it

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It shows that AF × ad = DF × CD is wrong
It should be AF × ad = DF × BD,
prove:
Let de be the center line on the hypotenuse of the right angle △ ACD, ﹥ de = AE = CE,
∴∠EAD=∠EDA,
﹤ DAF = ∠ BDF (plus 90 ° at the same time)
From ∠ is the common angle,
∴△DAF∽△BDF,
∴AF/DF=BD/AD.
Conclusion: AF × ad = BD × DF

As shown in the figure, ad is the height on the hypotenuse BC of RT △ ABC, and E is the midpoint of AC. the straight line ed intersects with the extension line of AB at point F. it is proved that the square of DF = af * BF

It is proved that: (a) ad ? BC, e is RT ? BC, e is RT ? BC, e is the midpoint of AC ᙨ AE = de = 1 / 2 · AC  DAE ∵ DAE = 1 / 2 · AC ᙨ DAE ? r t ⊙ ABC, ? ABC, ∵ BAC = 90 ∵ B ∵ C = 90 ° and RT △ ADC in the ADC ? DAE ? C = 90 ° and RT ᙨ ADC ? B 9 B DFIs: DF square

It is known that in △ ABC, ∠ C = 90 ° D is the midpoint of AB, e and F are on AC and BC respectively, and de ⊥ DF. Verification: AE2 + BF2 = ef2

It is proved that am ∥ BC is made through point a, and FD extension line is crossed at point M,
Connect em
∵AM∥BC，
∴∠MAE=∠ACB=90°，∠MAD=∠B．
∵AD=BD，∠ADM=∠BDF，
∴△ADM≌△BDF．
∴AM=BF，MD=DF．
And ∵ de ⊥ DF, ∵ EF = em
∴AE2+BF2=AE2+AM2=EM2=EF2．

In △ ABC, ∠ C = 90 ° D is the midpoint of AB, de ⊥ DF e is an arbitrary point on AC, f is an arbitrary point on BC In △ ABC, ∠ C = 90 ° D is the midpoint of AB, de ⊥ DF, e is any point on AC, f is an arbitrary point on BC Verification: EF square = AE square + BF square

Solution 1:
prove:
With D as the circle point and DB as the radius, rotate the DB side anticlockwise by 180 degrees, and then BD edge and ad edge coincide
Point B coincides with point A. mark that point F rotates to f '
∴AF'=FB,DF=DF',∠ADF'=∠FDB
∵ D is the midpoint of AB side,
Now point B and point a coincide
∵ RT △ ABC, ∵ C is the RT angle
Ψ A and ∠ B are mutually complementary
﹤ CAF '= 90 degrees
In RT △ AEF ', AE + f'b = EF'
∵ DF = DF ', and ∵ ADF' = ∠ FDB
ν ed vertical bisection FF '
∴EF=EF'
 AE + AF 'Square = AE + FB = EF' = EF '= EF
Proof of proposition
Solution 2:
Connect EF, de and DF as auxiliary line CD
It is proved that △ ade is all equal to △ CDF
Because ∠ ead = ∠ FCD = 45 ad = CD ∠ EDA = ∠ FDC
So △ ade is all equal to △ CDF
So AE = CF and CE = bf because AC = BC
Because the square of CF + the square of CE = the square of EF
So the square of AE + the square of BF = the square of EF

As shown in the figure, in RT △ ABC, ∠ C = 90 ° D is the midpoint of AB, e and F are on AC and BC respectively, and de ⊥ DF. It is proved that AE, EF and FB are the three sides of the same right triangle

It is proved that through point a, am ‖ BC is made, FD extension line is crossed with point m, and EM is connected
∵AM∥BC，
∴∠MAE=∠ACB=90°，∠MAD=∠B．
∵AD=BD，∠ADM=∠BDF，
∴△ADM≌△BDF．
∴AM=BF，MD=DF．
And ∵ de ⊥ DF, ∵ EF = em
∴AE2+BF2=AE2+AM2=EM2=EF2．
That is, AE, EF and FB are the three sides of the same right triangle

In the RT triangle ABC, ab = AC, angle a = 90 degrees, D is the point on BC, DF is perpendicular to AB, De is perpendicular to AC, M is the midpoint of BC Judge what triangle MEF is and prove it,

It is proved that if the crossing point m is mg ⊥ AC, MH ⊥ AB, and the vertical feet are g and h, then the quadrilateral agmh is a square ᚉ ah = mg
∠GMH=90°
 g, h are the midpoint of AC and ab  ah = 1 / 2Ab, CG = 1 / 2Ac = 1 / 2Ab  ah = CG ∵ m is the midpoint of BC ? mg = 1 / 2Ab, MH = 1 / 2Ac ∵ AB = AC ? mg = MH verifiable quadrilateral AEDF is a rectangle  de = AF verifiable ﹤ CDE is an isosceles triangle ? de = EC
∵ MHF = ∵ MGE = 90 ° mg = MH ∵ MFH ≌ △ Meg
﹤ MF = me ﹣ FMH = ∠ EMG ﹥ FMH + ∠ HME = ∠ EMG + ∠ HME, i.e., ∠ EMF = ∠ GMH = 90 degrees
△ MEF is an isosceles right triangle

In the RT triangle ABC, ab = AC, angle a = 90 ° point D is any point of BC, DF is perpendicular to AB and F, De is perpendicular to e, M is BC Those who want to come

The question is to judge the shape of △ MEF? If it is as follows:
Proof: link am
∵ BAC = 90 °, ab = AC, M is the midpoint of BC
∴AM =BM,∠BAM=∠CAM=45°,AM⊥BC
∵DF⊥AB,DE⊥AC,∠BAC=90°
ν the quadrilateral afde is a rectangle, ν DF = AE
∵DF⊥AB,∠B=45°,∴∠FDB=45°=∠B
∴BF=DF,∴BF=AE
In △ BFM and △ AEM
∴FM=EM,∠BMF=∠AME
∴AM⊥BC,∴∠BMF+∠AMF=90°
∴∠AME+∠AMF=∠EMF=90°
△ MEF is an isosceles right triangle