Question

multiple choice 10 points
from the information in the diagram, can you prove $\triangle fdg \cong \triangle fde$
diagram of a kite - shaped figure with triangles fdg and fde
\bigcirc yes, asa
\bigcirc yes, aaa
\bigcirc yes, sas
\bigcirc no

multiple choice 10 points
what additional information is needed to prove $\triangle mnp \cong \triangle pqm$ by sas?
diagram of a quadrilateral with triangles mnp and pqm
\bigcirc $\angle mpn \cong \angle mpq$
\bigcirc $\angle n \cong \angle q$
\bigcirc $\overline{pq} \cong \overline{mn}$

Explanation:

First Question:

Step1: Analyze triangle congruence criteria

We know the ASA (Angle - Side - Angle) criterion for triangle congruence states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, the triangles are congruent. For \(\triangle FDG\) and \(\triangle FDE\):

• \(\angle DGF\cong\angle DEF\) (marked angles)
• \(FD\) is a common side, so \(FD = FD\)
• \(\angle GDF\cong\angle EDF\) (marked angles)

So, by ASA, \(\triangle FDG\cong\triangle FDE\). AAA (Angle - Angle - Angle) only shows similarity, not congruence. SAS (Side - Angle - Side) requires two sides and the included angle, which is not the case here as we have two angles and the included side.

Step2: Eliminate incorrect options

• AAA is for similarity, not congruence, so "Yes, AAA" is wrong.
• SAS requires two sides and included angle, we have two angles and included side, so "Yes, SAS" is wrong.
• We can prove congruence, so "No" is wrong.

To prove \(\triangle MNP\cong\triangle PQM\) by SAS (Side - Angle - Side), we need two sides and the included angle.

• We know that \(MP\) is a common side, so \(MP = PM\).
• From the diagram, we can assume \(MN\) and \(PQ\) are related (maybe \(MN\parallel PQ\) or other properties), but for SAS, we need the included angle between the two sides. The sides around the angle would be \(MP\) and \(PN\) for \(\triangle MNP\) and \(MP\) and \(PQ\) for \(\triangle PQM\). The included angle would be \(\angle MPN\) for \(\triangle MNP\) and \(\angle MPQ\) for \(\triangle PQM\). So we need \(\angle MPN\cong\angle MPQ\).
• \(\angle N\cong\angle Q\) is an angle - angle - side or angle - side - angle related, not SAS.
• \(PQ\cong MN\) is a side - side - side or side - angle - side but not the included angle case for SAS here.

Answer:

A. Yes, ASA

Second Question: