In an isosceles right angle ABC with A 2,1 , B 2 , 4 , C m, n as vertices, the vertices C m,n may b

∵ab=ac,ad=ae,∠bac=∠dae.

abc=∠ade.(Two isosceles triangles with equal apex angles also have equal base angles) abc+ adf= ade+ adf=180°, get the points a, b, f, d on the same circle.

Therefore: afb= adb=90°, i.e., af bc; ab=ac

bf=fc.(The height on the base edge of the isosceles triangle is also the midline of the bottom edge).

The first: ab=ac, ad=ae, bac= dae

abc=∠ade.(Two isosceles triangles with equal apex angles also have equal base angles) abc+ adf= ade+ adf=180°, get the points a, b, f, d on the same circle.

Therefore: afb= adb=90°, i.e., af bc; ab=ac

bf=fc.(The height on the bottom edge of the isosceles triangle is also the middle line of the bottom edge) The second type: **The quantitative relationship between the line segment bf and cf, and explain the reason (1) Change abc and ade to equilateral triangle ab=ac, ad=ae, bac= dae

abc=∠ade.(Two isosceles triangles with equal apex angles.)

The equation for the straight line where ac is located is: y-2 = 3 (x+1), y = 3x + 3+2

ab|=-1-(-3)=2, isosceles triangle abc, so ab=ac or ba=bc

If ab=ac, then |ac|=2= (x 2+y 2)== ( 3 2+1)(x+1) 2,x=0,y= 3+2, then the equation for bc is: (y-2) [x-(-3)]=( 3+2-2) [0-(-3)],y= 3*x 3+ 3+2;

or x=-2, y=2-3, the equation for bc is: (y-2) [x-(-3)]=(2- 3-2) [0-(-3)], y=- 3*x 3- 3+2;

If ba=bc, then |bc|=|ab|=2, let the slope of bc be k, then the equation is y=k(x+3)+2, and there is an intersection with y= 3x+ 3+2, then k(x+3)+2= 3x+ 3+2, k=( 3x+ 3) (x+3) is substituted.

bc|= (k 2+1)*(x+3)=2: *x+3)=2

4x^2+12x+12)=2

x^2+3x+2=0

x = -1 (do not match, rounded) or x = -2, k = - 3, the BC equation is y=- 3 (x+3) + 2 = - 3x-3 3 + 2

The equation for ac is obtained from the point-oblique equation: kac 3 y-2 3 (x+1) is opened and reduced to a general formula. Then there is the two-point equation to find the equation for ab: we get x = 1.

Answer: (1) EP=FQ, prov: EAB=90°, EP AG, AG BC, EPA= EAB= AGB=90°, PEA+ EAP=90°, EAP+ BAG=90°, PEA= BAG, in EPA and AGB, EPA= BGA

pea=∠bag

ae=ab2) solution: eh=fh, the reason is: ep ag, fq ag, eph= fqh=90°, in eph and fqh, ehp= fhq

eph=∠fqh

ep=fq∴△eph≌△fqh（aas），eh=fh；

3）∵△eph≌△fqh，△epa≌△agb，△fqa≌△agc，∴s△fqas△agc，s△fqh=s△eph，s△epa=s△agb，s△aef=s△epa+s△fqa

s△agb+s△agc

s△abc=12

bc×ag=12

So the answer is: 288