in figure ABIIDE,AB=DE,ACIIBF and AC=DF prove that BCIIEF and BC=EF.
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fig in image...
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in triangle ABC & triangle DEF
AB=DE......GIVEN
ANGLE BAC= ANGLE EDF(ANGLES BETWEEN THE EQUAL SIDES OF TWO TRIANGLES ARE EQUAL)
AC=DF........GIVEN
BY SAS RULE TRIANGLE ABC & TRIANGLE DEF ARE CONGURENT
BY C.P.C.T
BC=EF
AB||DE &AC||DF
THEREFORE BC||EF AS THEY ARE EQUAL LIKE SIDES AB & DE;AND AC&DF
AB=DE......GIVEN
ANGLE BAC= ANGLE EDF(ANGLES BETWEEN THE EQUAL SIDES OF TWO TRIANGLES ARE EQUAL)
AC=DF........GIVEN
BY SAS RULE TRIANGLE ABC & TRIANGLE DEF ARE CONGURENT
BY C.P.C.T
BC=EF
AB||DE &AC||DF
THEREFORE BC||EF AS THEY ARE EQUAL LIKE SIDES AB & DE;AND AC&DF
SwapnilRokade:
And can you help me how it is parallel...
ok
thanks...
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