Known; a = 201x + 2011, B = 2010x + 2012, C = 2010x + 2013. Find a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ca

Known; a = 201x + 2011, B = 2010x + 2012, C = 2010x + 2013. Find a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ca

a-b=2010x+2011-(2010x+2012)=-1
a-c=2010x+2011-(2010x+2013)=-2
b-c=2010x+2012-(2010x+2013)=1
2a^2+2b^2+2c^2-2ab-2bc-2ca=（a-b)^2+(a-c)^2+(b-c)^2
=1+4+1=6
a^2+b^2+c^2-ab-bc-ca=6/2=3

It is known as follows: in △ ABC, the midpoint of AB, BC and Ca are e, F and g respectively, and ad is high

Proof: connect eg,
∵ e, F and G are the midpoint of AB, BC and Ca respectively,
ν EF is the median line of △ ABC, EF = 1
2AC．
(the median line of the triangle is half of the third side)
And ∵ ad ⊥ BC,
Ψ ADC = 90 ° and DG is the center line on the inclined side of right angle △ ADC,
∴DG=1
2AC．
(the center line on the hypotenuse of a right triangle is half of the hypotenuse)
∴DG=EF．
Similarly, de = FG, eg = Ge,
∴△EFG≌△GDE（SSS）．
∴∠EDG=∠EFG．

It is known as follows: in △ ABC, the midpoint of AB, BC and Ca are e, F and g respectively, and ad is high

Proof: connect eg,
∵ e, F and G are the midpoint of AB, BC and Ca respectively,
ν EF is the median line of △ ABC, EF = 1
2AC．
(the median line of the triangle is half of the third side)
And ∵ ad ⊥ BC,
Ψ ADC = 90 ° and DG is the center line on the inclined side of right angle △ ADC,
∴DG=1
2AC．
(the center line on the hypotenuse of a right triangle is half of the hypotenuse)
∴DG=EF．
Similarly, de = FG, eg = Ge,
∴△EFG≌△GDE（SSS）．
∴∠EDG=∠EFG．

In △ ABC, ah ⊥ BC is the midpoint of BC, Ca and ab at h, D, e and f (as shown in the figure). It is proved that ﹣ def = ∠ HFE

It is proved that ∵ E and F are the midpoint of AC and ab respectively,
∴EF∥BC，
According to the parallel line theorem, ∠ HFE = ∠ FHB, ∠ def = ∠ CDE;
Similarly, it can be proved that ∠ CDE = ∠ B,
∴∠DEF=∠B．
And ∵ ah ⊥ BC, and F is the midpoint of ab,
∴HF=BF，
∴∠B=∠BHF，
∴∠HFE=∠B=∠DEF．
That is ∠ HFE = ∠ def

As shown in the figure, △ ABC, e, F, G are the midpoint of AB, BC and Ca respectively, and ad is high. It is proved that: ∠ EDG = ∠ EFG

Parallelogram aefg
EFG=EAG
ADB=90 AE=BE
DE=EA EAD=EDA
Similarly, Ag = DG DAG = Gda
EAG=EDG=EFG

As shown in the figure, in the triangle ABC, ad ⊥ BC is on D, e is a point on AB, EF ⊥ BC is at F, DG ∥ Ba is intersected with Ca in G, and it is proved that ∠ 1 = ∠ 2

prove:
∵DG∥BA
Ψ 1 = ∠ 3 (equal internal staggered angle)
∵AD⊥BC,EF⊥BC
∴EF∥AD
Ψ 2 = ∠ 3 (equal position angle)
∴∠1＝∠2

D. E and G are the points on △ ABC, BC, CA, AB respectively, DG ∥ AC, DG = CE, extend eg to F, make ef = 2EG, connect CF, try to explain: CF and DG are equally divided

Proof: connect DF and CG
∵DG∥AC,DG=CE
﹣ parallelogram CDGE (opposite sides parallel and equal)
∴EF∥BC,EG＝CD
∵EF＝EG+FG,EF＝2EG
∴FG＝EG
∴FG＝CD
The parallelogram CDFG (opposite sides parallel and equal)
The CF and DG are equally divided

It is known that: in △ ABC, ab = AC, draw an arc with point a as the center of the circle, respectively intersect the extension line ab of Ca and E.F, connect EF and extend the intersection BC to g, and verify: eg ⊥ BC

prove;
∵ e, f is the intersection point of the arc drawn with the extension line of Ca and ab with point a as the center of the circle
∴AE=AF
∴∠E=∠AFE=∠BFG
∵AB=AC
∴∠B=∠C
∴∠B+∠BFG=∠C+∠E
∴∠EGC=∠FGB=180º÷2=90º
Eg ⊥ BC

In △ ABC, the point DEF is on bcab AC, BD = CF, be = CD, ab = AC, DG ⊥ EF is at point G, and eg = FG is proved

Proof: because AB = AC, so ∠ B = ∠ C
BD = CF, be = CD
So △ BDE ≌ △ CFD
Then de = DF
So △ DEF is an isosceles triangle,
DG ⊥ ef
So eg = FG

It is known that in △ ABC, the angles a = 90 degrees, D, F, e are the midpoint of BC, CA, AB, respectively Please answer in detail

prove:
∵ D is the midpoint of BC, ∵ a = 90 ᙽ
/ / ad = 1 / 2BC (the center line of the hypotenuse of a right triangle is equal to half of the hypotenuse)
∵ E and F are the midlines of AB and AC respectively
The EF is the median line of △ ABC
∴EF=1/2BC
∴AD =EF