Question

In triangle $ABC$and $DEF$, $AB = FD$ and $\angle A = \angle D$. The two
triangle will be congruent by SAS axiom if
A. $BC = EF$
B. $AC = DE$
C. $AC = EF$
D. $BC = DE$

Answer 1

Hint: In this problem, we have to check for each side of the given two triangle $ABC$ and
$DEF$ so that both the triangles are congruent by satisfying the SAS axiom. It is given that $AB =
FD$ and $\angle A = \angle D$. So, we need to equate two corresponding sides of the triangle
$\Delta ABC$ and $\Delta DEF$.

For the congruency of two triangles it must follow the SAS axiom.
For $\Delta ABC$ and $\Delta DEF$ it is given that $AB = FD$ and $\angle A = \angle D$.
Now considering $BC = EF$
For $\Delta ABC$ and $\Delta DEF$,
$AB = FD$ (given)
$\angle A = \angle D$ (given)
$BC = EF$
$\therefore \Delta ABC \cong \Delta DEF$
But in this case $\Delta ABC$ and $\Delta DEF$ follows SSA axioms.
Thus, when $BC = EF$ the triangle did not satisfy the SAS axiom.
Considering $AC = DE$
For $\Delta ABC$ and $\Delta DEF$,
$AB = FD$ (given)

$\angle A = \angle D$ (given)
$AC = DE$
$\therefore \Delta ABC \cong \Delta DEF$
Therefore, when $AC = DE$triangle $\Delta ABC$ and $\Delta DEF$ congruent by SAS
axioms.
Considering $AC = EF$
For $\Delta ABC$ and $\Delta DEF$
$AB = FD$ (given)
$\angle A = \angle D$ (given)
$AC = EF$
$\therefore \Delta ABC \cong \Delta DEF$
Here $\Delta ABC$ and $\Delta DEF$ follows SSA axioms.
Thus, when $AC = EF$ the triangle did not satisfy the SAS axiom.

Considering $BC = DE$
For $\Delta ABC$ and $\Delta DEF$
$AB = FD$ (given)
$\angle A = \angle D$ (given)
$BC = DE$
$\therefore \Delta ABC \cong \Delta DEF$
$\Delta ABC$and$\Delta DEF$ follows SSA axioms.
Thus, when $BC = DE$ the triangle did not satisfy the SAS axiom.

Hence, the correct option is B.
Note: Here, we have to find the correct condition so that the triangle $\Delta ABC$ and $\Delta
DEF$ becomes congruent by following the SAS axioms. Since the side $AB = FD$ and $\angle A
= \angle D$. From that data we can easily find out the required equal sides of the triangles for
which both the triangle $\Delta ABC$ and $\Delta DEF$ become congruent by SAS axioms.