Question

triangle congruence: asa and aas
given: ∠tsr and ∠qrs are right angles; ∠t ≅ ∠q
prove: △tsr ≅ △qrs
step 1: we know that ∠tsr ≅ ∠qrs because all right angles are congruent.
step 2: we know that ∠t ≅ ∠q because it is given.
step 3: we know that sr ≅ rs because of the reflexive property.
step 4: △tsr ≅ △qrs because
cards: of the asa congruence theorem., of the third angle theorem., of the aas congruence theorem., all right triangles are congruent.

Explanation:

Step 1: Analyze Step 1's Missing Part

We know that \( \angle TSR \cong \angle QRS \) because all right angles are congruent. So the missing part for Step 1 is "all right angles are congruent".

Step 2: Analyze Step 2's Missing Part

We know that \( \angle T \cong \angle Q \) (given), and we already have a right angle congruent. For the third angle, by the third angle theorem, the remaining angles will be congruent. Wait, no, let's check the congruence. Wait, we have two angles: right angle and \( \angle T \cong \angle Q \), and the side \( SR \cong RS \) (reflexive). So the congruence theorem here: AAS (two angles and a non - included side) or ASA? Wait, let's list the parts. \( \angle TSR \) (right angle) \( \cong \angle QRS \) (right angle), \( \angle T \cong \angle Q \), and \( SR = RS \) (common side). So this is AAS? Wait, no, AAS is two angles and a non - included side, ASA is two angles and the included side. Wait, the side \( SR \) is between \( \angle T \) and \( \angle TSR \) in \( \triangle TSR \), and between \( \angle Q \) and \( \angle QRS \) in \( \triangle QRS \)? Wait, no, \( SR \) is a leg of both right triangles. Wait, \( \angle T \), \( \angle TSR \), and side \( SR \); \( \angle Q \), \( \angle QRS \), and side \( RS \) (which is equal to \( SR \)). So actually, it's AAS? Wait, no, let's re - examine. The AAS congruence theorem states that if two angles and a non - included side of one triangle are congruent to the corresponding two angles and non - included side of another triangle, then the triangles are congruent. Here, we have \( \angle T \cong \angle Q \), \( \angle TSR \cong \angle QRS \), and \( SR \cong RS \) (the side is not included between the two angles? Wait, \( \angle T \) and \( \angle TSR \) have side \( TR \) included, but \( SR \) is another side. Wait, maybe I made a mistake. Wait, the right angle, \( \angle T \cong \angle Q \), and the common side. So the congruence theorem here is AAS? Wait, no, let's check the options. The options for the last step (Step 4) are "of the ASA congruence theorem", "of the third angle theorem", "of the AAS congruence theorem", and "all right angles are congruent". Wait, Step 4: We have two angles (\( \angle T \cong \angle Q \), \( \angle TSR \cong \angle QRS \)) and a side (\( SR \cong RS \)). So this is AAS (two angles and a non - included side) or ASA? Wait, in \( \triangle TSR \) and \( \triangle QRS \): \( \angle T \cong \angle Q \), \( \angle TSR \cong \angle QRS \), \( SR \cong RS \). The side \( SR \) is opposite to \( \angle T \) in \( \triangle TSR \) and opposite to \( \angle Q \) in \( \triangle QRS \)? No, \( SR \) is adjacent to \( \angle TSR \) and \( \angle T \). Wait, maybe it's AAS. So the missing part for Step 4 is "of the AAS congruence theorem". But let's go step by step.

Wait, the first missing part (Step 1) is "all right angles are congruent" (the bottom - right card). Then Step 2: "of the third angle theorem"? No, Step 2 is about \( \angle T \cong \angle Q \) (given), and then Step 3 is \( SR \cong RS \) (reflexive). Then Step 4: we have two angles (\( \angle T \), \( \angle TSR \)) and a side (\( SR \)) congruent to two angles (\( \angle Q \), \( \angle QRS \)) and a side (\( RS \)). So by AAS congruence theorem. So the card for Step 4 is "of the AAS congruence theorem" (the bottom - left card).

Wait, let's match the cards:

- Step 1: "all right angles are congruent" (bottom - right card)
- Step 2: "of the third angle theorem"? No, Step 2 is just stating \( \angle T \cong \angle Q \) is given, maybe the card for Step 2 is not needed? Wait, the problem is to match the steps with the cards. Let's list the steps:

Step 1: We know that \( \angle TSR \cong \angle QRS \) because [all right angles are congruent] (bottom - right card).

Step 2: We know that \( \angle T \cong \angle Q \) because it is given.

Step 3: We know that \( SR \cong RS \) because of the reflexive property.

Step 4: \( \triangle TSR \cong \triangle QRS \) because [of the AAS congruence theorem] (bottom - left card). Wait, no, AAS is two angles and a non - included side. We have two angles (\( \angle T \), \( \angle TSR \)) and a side (\( SR \)) which is a non - included side? Wait, \( \angle T \) and \( \angle TSR \) have included side \( TR \), but \( SR \) is a leg. Alternatively, maybe it's ASA? No, ASA requires the included side. Wait, the right angle, \( \angle T \cong \angle Q \), and the side \( SR \) which is between \( \angle T \) and \( \angle TSR \)? No, \( SR \) is perpendicular to \( TS \) and \( RQ \)? Wait, the diagram shows \( \angle TSR \) and \( \angle QRS \) are right angles, so \( TS \perp SR \) and \( RQ \perp SR \), so \( TS \parallel RQ \). Then \( \angle T \) and \( \angle Q \) are alternate interior angles? Maybe not. Let's get back to the congruence. We have:

- \( \angle T \cong \angle Q \) (given)
- \( \angle TSR \cong \angle QRS = 90^{\circ}\) (all right angles are congruent)
- \( SR \cong RS \) (reflexive property)

So in \( \triangle TSR \) and \( \triangle QRS \), we have two angles and a non - included side congruent (AAS: \( \angle T \), \( \angle TSR \), and \( SR \); \( \angle Q \), \( \angle QRS \), and \( RS \)). So the congruence theorem is AAS. Therefore, Step 4 uses the AAS congruence theorem.

So the missing parts:

Step 1: all right angles are congruent (bottom - right card)

Step 4: of the AAS congruence theorem (bottom - left card)

But maybe the question is to fill in the last step's reason. So the answer for the last step (Step 4) is "of the AAS congruence theorem".

Answer:

of the AAS congruence theorem