Question

determine if the triangles are congruent. explain your reasoning.
1.
triangle edf with side ed=8 mm, angle at e=25°, angle at f=45°; triangle abc with side bc=8 mm, angle at b=25°, angle at c=45°
2.
right triangle qsr with leg qs=4 in., right angle at s; right triangle utv with leg tv=4 in., right angle at v
3.
isosceles triangle hjk with hj=hk=2.2 ft, angles at l and k=63°, and a segment hj
4.
quadrilateral wxyz with wz=xy=5 m, and triangle xyz

1. write a two - column proof.

given: $overline{ac}$ bisects $\angle bad$, $\angle bca \cong \angle dca$
prove: $bc = dc$
triangle abc and adc with ac as a common side, and a right angle at c

statements	reasons
2.	2.
3.	3.
4.	4.
5.	5.
6.	6.
7.	7.

Explanation:

Problem 1:

Step1: Identify angles and sides

In $\triangle DEF$ and $\triangle ABC$ (wait, labels: $\triangle DEF$: $\angle E = 25^\circ$, $\angle F = 45^\circ$, side $DE$? Wait, no, $\triangle DEF$: side $EF = 8$ mm? Wait, $\triangle DEF$: $\angle E = 25^\circ$, $\angle F = 45^\circ$, side between? Wait, $\triangle ABC$ (labels: $A, B, C$: $\angle B = 25^\circ$, $\angle C = 45^\circ$, side $BC = 8$ mm. $\triangle DEF$: $\angle E = 25^\circ$, $\angle F = 45^\circ$, side $EF = 8$ mm? Wait, angle - side - angle (ASA)? Wait, $\angle E = 25^\circ$, side $EF = 8$ mm, $\angle F = 45^\circ$; in $\triangle ABC$: $\angle B = 25^\circ$, side $BC = 8$ mm, $\angle C = 45^\circ$. So by ASA congruence (two angles and included side), the triangles are congruent.

Step2: Apply ASA criterion

ASA 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. Here, $\angle E \cong \angle B = 25^\circ$, side $EF \cong BC = 8$ mm, $\angle F \cong \angle C = 45^\circ$. So $\triangle DEF \cong \triangle BAC$ (or $\triangle DFE \cong \triangle BCA$) by ASA.

Step1: Identify right angles and sides

$\triangle QSR$ is a right triangle at $S$, with $QS = 4$ in, right angle at $S$. $\triangle TUV$ is a right triangle at $V$, with $TV = 4$ in, right angle at $V$. Wait, but the angles: $\triangle QSR$: right angle at $S$, $\angle Q$; $\triangle TUV$: right angle at $V$, $\angle T$. Wait, are the sides and angles corresponding? Wait, maybe AAS or HL? Wait, $\triangle QSR$: right angle at $S$, leg $QS = 4$ in; $\triangle TUV$: right angle at $V$, leg $TV = 4$ in. But the other angles: $\angle Q$ and $\angle T$? Wait, maybe not. Wait, no, maybe the triangles are congruent by AAS? Wait, $\triangle QSR$: $\angle S = 90^\circ$, $QS = 4$ in, $\angle Q$; $\triangle TUV$: $\angle V = 90^\circ$, $TV = 4$ in, $\angle T$. Wait, maybe the triangles are congruent by HL? No, HL is for hypotenuse - leg. Wait, maybe the triangles are congruent by AAS: $\angle S = \angle V = 90^\circ$, $QS = TV = 4$ in, and $\angle Q = \angle U$? Wait, no, maybe the triangles are congruent by ASA? Wait, no, let's check: $\triangle QSR$: right angle at $S$, $QS = 4$ in, $\angle Q$; $\triangle TUV$: right angle at $V$, $TV = 4$ in, $\angle T$. Wait, maybe the triangles are congruent by AAS because $\angle S=\angle V = 90^\circ$, $QS = TV = 4$ in, and $\angle Q=\angle U$ (if we consider the other angles). Wait, alternatively, maybe the triangles are congruent by SAS? No, right angle, leg, and angle. Wait, actually, $\triangle QSR$: $\angle S = 90^\circ$, $QS = 4$ in, $\angle Q$; $\triangle TUV$: $\angle V = 90^\circ$, $TV = 4$ in, $\angle T$. If we rotate one triangle, we can see that $\triangle QSR \cong \triangle TUV$ by AAS (two angles and a non - included side) or ASA? Wait, maybe the triangles are congruent because $\angle S=\angle V = 90^\circ$, $QS = TV = 4$ in, and $\angle Q=\angle T$? Wait, no, maybe the triangles are congruent by HL? No, HL requires hypotenuse and leg. Wait, maybe the triangles are congruent by AAS: $\angle S=\angle V = 90^\circ$, $QS = TV = 4$ in, and $\angle R=\angle U$? Wait, perhaps the key is that they are right triangles with one leg equal and one acute angle equal, so by AAS, they are congruent.

Step2: Apply AAS criterion

AAS states that if two angles and a non - included side of one triangle are congruent to two angles and the corresponding non - included side of another triangle, the triangles are congruent. Here, $\angle S=\angle V = 90^\circ$, $QS = TV = 4$ in, and $\angle Q=\angle T$ (or $\angle R=\angle U$), so by AAS, $\triangle QSR \cong \triangle TUV$.

Step1: Identify sides and angles

In $\triangle HLJ$ and $\triangle HKJ$: $HL = HK = 2.2$ ft, $\angle L=\angle K = 63^\circ$, and $HJ$ is common. So by SAS (Side - Angle - Side) or ASA? Wait, $HL = HK$, $\angle L=\angle K$, and $LJ$ and $KJ$? Wait, $HJ$ is the common side. So $\angle L=\angle K = 63^\circ$, $HL = HK = 2.2$ ft, and $\angle HLJ=\angle HKJ$? Wait, no, $\triangle HLJ$ and $\triangle HKJ$: $HL = HK$, $\angle L=\angle K$, $HJ = HJ$ (common side). So by SAS (side - angle - side: $HL = HK$, $\angle L=\angle K$, $LJ$ and $KJ$? Wait, no, $HJ$ is the side between? Wait, $\angle L$ and $\angle K$ are at $L$ and $K$, with $HL = HK$, and $HJ$ is the altitude. So $\triangle HLJ \cong \triangle HKJ$ by SAS: $HL = HK$, $\angle L=\angle K$, $LJ = KJ$? Wait, no, $HJ$ is common. Wait, actually, $HL = HK$, $\angle L=\angle K$, and $HJ$ is common, so by SAS (side - angle - side: two sides and included angle? Wait, $\angle L$ is between $HL$ and $LJ$, $\angle K$ is between $HK$ and $KJ$. But $HL = HK$, $\angle L=\angle K$, and $HJ$ is common. So maybe by AAS? Wait, no, $HL = HK$, $\angle L=\angle K$, $HJ = HJ$, so by SAS (if we consider $HL$, $\angle L$, $LJ$ and $HK$, $\angle K$, $KJ$; but $LJ = KJ$? Wait, $HJ$ is the altitude, so $LJ = KJ$ (since $HL = HK$ and $\angle L=\angle K$, the triangle is isoceles, so $HJ$ bisects $LK$). So $\triangle HLJ \cong \triangle HKJ$ by SAS: $HL = HK$, $\angle L=\angle K$, $LJ = KJ$ (or by SAS with $HL$, $\angle L$, $HJ$ and $HK$, $\angle K$, $HJ$? Wait, $HJ$ is common, so $HL = HK$, $\angle L=\angle K$, $HJ = HJ$, so by SAS (side - angle - side: two sides and included angle. The included angle for $HL$ and $HJ$ is $\angle HJL$, and for $HK$ and $HJ$ is $\angle HJK$. But since $\angle L=\angle K$ and $HL = HK$, $\angle HJL=\angle HJK = 90^\circ$? Wait, maybe by SAS: $HL = HK$, $\angle L=\angle K$, $LJ = KJ$ (since $HJ$ is the altitude, it bisects $LK$), so $\triangle HLJ \cong \triangle HKJ$ by SAS.

Step2: Apply SAS criterion

SAS states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, the triangles are congruent. Here, $HL = HK$, $\angle L=\angle K$, and $LJ = KJ$ (or $HJ$ is common, and $\angle L$ and $\angle K$ are the included angles with $HL$ and $HK$ respectively), so $\triangle HLJ \cong \triangle HKJ$.

Answer:

The triangles are congruent by ASA (Angle - Side - Angle) congruence criterion, as two angles and the included side of one triangle are congruent to two angles and the included side of the other triangle.

Problem 2: