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2016-03-18T19:41:10+05:30

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GIVEN: A llgm ABCD
TO PROVE : A  DIAGONAL AC OF llgm ABCD DIVIDES IT INTO 2 CONGRUENT TRIANGLES  ABC AND CDA 
CONSTRUCTION : JOIN AC
PROOF : SINCE ABCD IS A llgm 
     AB ll DC AND AD ll BC
NOW , AD ll BC AND TRANSVERSAL AC INTERSECTS THEM AT A AND C 
     => <DAC = <BCA              --- 1 [ALTERNATE INTERIOR ANGLES]
AGAIN , AB ll DC AND TRANSVERSAL AC INTERSECTS THEM AT A AND C RESPECTIVELY
     =>  <BAC = <DCA             ---- 2 [ALTERNATE INTERIOR ANGLES ]
IN TRIANGLES ABC AND CDA , 
<BCA  =  <DAC  [FROM 1]
AC  =  AC    [COMMON SIDE]
<BAC  =  <DCA   [FROM 2 ]

TRIANGLES ARE CONGRUENT 
                            HENCE THE PROOF


AD = CB  AND  DC = BA 
[SINCE CORRESPONDING PARTS OF CONGRUENT TRIANGLES ARE EQUAL]
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