User:Mohammed farag

THEORIES DISCOVERED BY MOHAMMED FARAG:[edit]

(1)

in any triangle ABC

where a,b,c are the lengths of its sides opposing the angles A,B,C in sequence:

PROVE:

sinA=sin(B+C)

sinA=sinBcosC+cosBsinC

 1 =cosC*sinB/sinA+cosB*sinC/sinA

 1 =cosC*b/a+cosB*c/a

 a =bcosC+ccosB

(2)

ABCD QUADRANGULAR AREA=AC*BD*sinAMB/2

where M is the point of decussation of AC ,&BD

PROVE:

IN A QUADRANGLE ABCD WITH M POINT IS THE CROSS POINT OF AC &BD

THE QUADRANGULAR AREA=THE SUM OF ALL ITS PARTS AREAS

       =TRIANGLE AMB AREA + TRIANGLE BMC AREA + TRIANGLE CMD AREA +TRIANGLE DMA AREA

       =AM*MB*SIN(AMB)/2 + BM*MC*SIN(BMC)/2 + CM*MD*SIN(CMD)/2 + DM*MA*SIN(DMA)/2

       =(AM*MB+MB*MC+MD*MC+MD*MA)*SIN(AMB)/2

       =(MB(AM+MC)+MD(AM+MC))*SIN(AMB)/2
     
       =(MB+MD)(AM+MC)*SIN(AMB)/2

       =AC*BD*SIN(AMB)/2

(3)

In ABCD quadrangle

where EI cuts AB,BC,CD,DA in F,E,H,I in sequence:

AF*BE*CH*DI/(FB*EC*HD*IA)=1

PROVE:

IN THE FIGURE DESCRIBED IF WE DRAW PERPENDICULAR LINES FROM A,B,C,D POINTS TO EI LINE IN J,K,L,M CONSECUTIVELY

SO,FROM THE LAWS OF PARALLELS WE CAN GET THE FOLLOWINGS:

AF/FB=AJ/BK---------1

BE/EC=BK/CL---------2

CH/HD=CL/DM---------3

DI/IA=DM/AJ---------4

BY MULTIPLYING THE LEFT SIDES & THE RIGHT SIDES OF THE ABOVE FOUR EQUATIONS WE CAN GET:

(AF/FB)*(BE/EC)*(CH/HD)*(DI/IA)=(AJ/BK)*(BK/CL)*(CL/DM)*(DM/AJ)

 SO,AF*BE*CH*DI/(FB*EC*HD*IA)=1

 This theory can be applied also on pentangulars,hexangulars,.....etc

(4)

Triangular area =a*b*c/4r

where a,b,c are the lengths of sides opposing the angles in sequence & where r is the radius of the circle passing by all points of the triangle ABC

PROVE:

TRIANGULAR AREA =a*b*sin(c)/2

               =a*b*c/4r

(5)

In ABC triangle :

sinA*sinB*sinC=cosB*sin(A+C)+cosA*sinB*cosC

(6)

In a triangle ABC :

sinA*cosA=(b^2+c^2-a^2)*a/4rbc

where r is the radius of the circle passing all the points of the triangle ABC

PROVE:

COS(A)=(b^2+c^2-a^2)/2c.b ------ * a.2r/a.2r where r is the radius of the circle passing all the points of the triangle ABC

SO, SIN(A)*COS(A)=a(b^2+c^2-a^2)/4r*b*c

where a,b,c are the lengths of sides opposing the angles in sequence

(7)

In same conditions of the triangle in condition(6)

tan(A)=abc/r(b^2+c^2-a^2)

PROVE:

tan(A)= SIN(A)/COS(B)

     = 2.a.b.c/2r(b^2+c^2-a^2)

So, tan(A) = a.b.c/r(b^2+c^2-a^2)

(8)

In any quadrangle ABCD :

sinDAC*sinABD*sinBCA*sinCDB/(sinCAB*sinDBC*sinACD*sinBDA)=1 PROVE:

IN ANY QUADRANGLE ABCD WHERE X IS THE CROSS POINT OF AC & BD BY THE LAW OF SINES APPLIED ON THE FOLLOWING TRIANGLES:

TRIANGLE XAD----- XA/XD=SIN(ADB)/SIN(CAD)

TRIANGLE XAB----- XB/XA=SIN(CAB)/SIN(ABD)

TRIANGLE XBC----- XC/XB=SIN(DBC)/SIN(ACB)

TRIANGLE XCD----- XD/XC=SIN(ACD)/SIN(CDB)

BY MULTIPLYING EACH SIDE OF THE FOUR EQUATIONS ABOVE WE CAN GET THE THEORY

This theory can be applied on pentangles,hexangles.....etc

(9)

In any pyramid with apex X & base ABCD :

sin XAB*sinXBC*sinXCD*sinXDA/(sinXBA*sinXCB*sinXDC*sinXAD)=1 prove:

     in triangle XAB XA/AB=SIN(XBA)/SIN(XAB)
     IN TRIANGLE XBC XB/XC=SIN(XCB)/SIN(XBC)
     IN TRIANGLE XCD XC/XD=SIN(XDC)/SIN(XCD)
     IN TRIANGLE XDA XD/XA=SIN(XAD)/SIN(XDA)

BY MULTIPLYING EACH SIDE OF THE ABOVE FOUR EQUATIONS WE CAN GET THE LAW

This theory can be applied on pyramids with (quadrangular ,pentangular,hexangular....etc)bases

(10)

THE PENTAGONAL STAR THEORY OF MOHAMED FARAG

IN THE PENTAGON ABCDE WHERE ALL POINTS CONNECTED TOGETHTER BY INTERIOR LINES WHICH MAKE A STAR SHAPE WHERE:

F POINT IS THE DECUSSATION POINT OF AC & BE

G POINT IS THE DECUSSATION POINT OF AC & BD

H POINT IS THE DECUSSATION POINT OF BD & CE

I POINT IS THE DECUSSATION POINT OF AD & CE

J POINT IS THE DECUSSATION POINT OF AD & BE

(JA/AF)*(FB/BG)*(GC/CH)*(HD/DI)*(IE/EJ)=1

PROVE:

IN THE TRIANGLE JAF, AJ/AF=SIN(AFJ)/SIN(AJF) 1

IN THE TRIANGLE BFG, BF/BG=(SIN(BGF)/SIN(BFG) 2

IN THE TRIANGLE CHG, CG/CH =SIN(CHG)/SIN(CGH) 3

IN THE TRIANGLE HDI, DH/DI =SIN(DIH)/SIN(DHI) 4

IN THE TRIANGLE EIJ, EI/EJ =SIN(EIJ)/SIN(EJI) 5

BY MULTIPLYING THE LEFT SIDES & THE RIGHT SIDES OF THE ABOVE FIVE EQUATIONS,

WE CAN GET THE FOLLOWING:

(JA/AF)*(FB/BG)*(GC/CH)*(HD/DI)*(IE/EJ) = (SIN(AFJ)/ SIN(AJF))*((SIN(BGF)/SIN(BFG))*(SIN(CHG)/SIN(CGH))*(SIN(DIH)/SIN(DHI))*(SIN(EIJ)/SIN(EJI))

BUT THROUGH DECUSSATIONS OF THE STRAIGHT INTERIOR LINES BETWEEN THE POINTS OF THE PENTAGON:

SIN(AFJ)=SIN(BFG)

,SIN(AJF)=SIN(IJE)

,SIN(CGH)=SIN(BGF)

,SIN(CHG)=SIN(DHI)

, & SIN(DIH)=SIN(EIJ)

SO,

(SIN(AFJ)/ SIN(AJF))*((SIN(BGF)/SIN(BFG))*(SIN(CHG)/SIN(CGH))*(SIN(DIH)/SIN(DHI))*(SIN(EIJ)/SIN(EJI))=1

SO,

(JA/AF)*(FB/BG)*(GC/CH)*(HD/DI)*(IE/EJ)=1

THESE THEORIES WERE DISCOVERED BY MOHAMMED FARAG

EGYPTMOHAMMED@YAHOO.COM

M.FARAG59@YAHOO.COM

LOSTING_LOVER@HOTMAIL.COM

EGYMOHA@GMAIL.COM

0020827663354

0166821164--Mohammed farag (talk) 20:48, 24 April 2008 (UTC)