In triangles ABC and DEF,AB = FD and angle A | vedclass

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(D) According to the $SAS$ (Side-Angle-Side) congruence axiom,two triangles are congruent if two sides and the included angle of one triangle are equa

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$AC = DE$

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(D) According to the $SAS$ (Side-Angle-Side) congruence axiom,two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle.Given: $AB = FD$ and $\angle A = \angle D$.For the triangles to be congruent by $SAS$,the sides forming the angle $\angle A$ (which are $AB$ and $AC$) must be equal to the sides forming the angle $\angle D$ (which are $FD$ and $DE$).Since $AB = FD$ is already given,we must have $AC = DE$ for the $SAS$ condition to be satisfied.

In triangles $ABC$ and $DEF$,$AB = FD$ and $\angle A = \angle D$. The two triangles will be congruent by the $SAS$ axiom if:

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