Question

Karen notices that segment BC and segment EF are congruent in the image below:

Answer 1

Answer:

Givens:

• $A(-2;2) \ B(-2;4) \ C(0;3) \ D(2;1) \ E(2;3) \ F(0;2)$
• $BC \cong EF$
• $\angle C \cong \angle F$

Using this information we could recur to SAS postulate to demonstrate the congruence of this pair of triangles. We already have one side and one angle congruent, we just need to demonstrate that $AC \cong DF$, that will prove the complete congruence.

So, based on the given coordinates, we're able to calculate the length of each side and find if they are the same. We use this formula:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\d_{AC} =\sqrt{(0-(-2))^{2}+(3-2)^{2}}=\sqrt{4+1}=\sqrt{5}\\\\\d_{DF} =\sqrt{(0-2)^{2}+(2-1)^{2}}=\sqrt{4+1}=\sqrt{5}$

Therefore, $AC \cong DF$, because they have the same length.

Now, based on these congruences:

• $BC \cong EF$
• $\angle C \cong \angle F$
• $AC \cong DF$

We demonstrate by the Side-Angle-Side (SAS) postulate that ΔABC ≅ ΔDEF.