Below are \triangle ABC△ and \triangle DEF△DEFtriangle, D, E, F. We assume that AB=DEAB=DEA, B, equals, D, E, BC=EFBC=EFB, C, equals, E, F, and AC=DFAC=DFA, C, equals, D, F.

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Step-by-step explanation:

Justification of step 1 is based on the congruence of segments AB and DE.

Therefore correct choice is the first one:

A) AB = DE and segments with same length are congruent

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

hope it

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )