Question

2 determine the information that is needed to use the indicated theorem to show that the triangles are congruent.
a) δfjg ≅ δhjg by sas
b) δvwx ≅ δzyx by asa
c) δkjl ≅ δkts by sas
d) δdef ≅ δdgf by asa

Explanation:

Part (a)

Step1: Recall SAS Congruence

SAS (Side - Angle - Side) congruence 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.
In \(\triangle FJG\) and \(\triangle HJG\), we know that \(JG\) is common (so \(JG\cong JG\)) and \(\angle FJG=\angle HJG = 90^{\circ}\) (right angles). For SAS, we need the other pair of sides adjacent to the right angle to be congruent. So we need \(FJ\cong HJ\).

Step2: Verify the Requirement

To apply SAS to \(\triangle FJG\cong\triangle HJG\), we have one side \(JG\) common, one angle \(\angle FJG=\angle HJG\) (right angles), and we need the other side forming the angle (the side from \(J\) to \(F\) and \(J\) to \(H\)) to be congruent, i.e., \(FJ = HJ\) (or \(FJ\cong HJ\)).

Part (b)

Step1: Recall ASA Congruence

ASA (Angle - Side - Angle) 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.
In \(\triangle VWX\) and \(\triangle ZYX\), we know that \(VX\cong ZY\) (marked with tick marks) and \(\angle VXW=\angle ZXY\) (vertical angles, so they are congruent). For ASA, we need another pair of angles (the angles at \(W\) and \(Z\) or at \(V\) and \(Y\)) to be congruent. Specifically, we need \(\angle WVX\cong\angle ZYX\) (or \(\angle VWX\cong\angle ZYX\) and \(\angle V\cong\angle Z\) depending on the angle - side - angle order). More precisely, we need \(\angle VWX\cong\angle ZYX\) (the angles at \(W\) and \(Y\)) or \(\angle V\cong\angle Z\) (the angles at \(V\) and \(Z\)) along with the included side \(VX\cong ZY\) and the vertical angle \(\angle VXW=\angle ZXY\). The key missing piece for ASA is that we need \(\angle VWX\cong\angle ZYX\) (or \(\angle V\cong\angle Z\)) and the included side. But from the diagram, the vertical angles are equal, the side \(VX\cong ZY\), so we need \(\angle VWX\cong\angle ZYX\) (the angles at \(W\) and \(Y\)) or \(\angle V\cong\angle Z\) (the angles at \(V\) and \(Z\)). The most direct is that we need \(\angle VWX\cong\angle ZYX\) (or \(\angle V\cong\angle Z\)) to apply ASA. In simpler terms, we need \(\angle W\cong\angle Y\) (or \(\angle V\cong\angle Z\)) and the included side \(VX\cong ZY\) and the vertical angle \(\angle VXW=\angle ZXY\). So the missing information is that \(\angle VWX\cong\angle ZYX\) (or \(\angle V\cong\angle Z\)).

Part (c)

Step1: Recall SAS Congruence

For SAS congruence, we need two sides and the included angle. In \(\triangle KJL\) and \(\triangle KTS\), we know that \(KJ\) and \(KT\) are parts of the same line, \(LJ\perp JT\) and \(ST\perp JT\) (right angles), and \(LJ\cong ST\) (marked with tick marks). We also know that \(\angle KJL=\angle KTS = 90^{\circ}\). For SAS, we need the other pair of sides adjacent to the right angle, i.e., \(KJ\cong KT\) (or the side from \(K\) to \(J\) and \(K\) to \(T\)) to be congruent, and also the side \(KL\) and \(KS\)? No, wait, the triangles are \(\triangle KJL\) and \(\triangle KTS\). So sides: \(LJ\cong ST\) (given by tick marks), \(\angle KJL=\angle KTS = 90^{\circ}\) (right angles), and we need \(KJ\cong KT\) (the other leg of the right triangle) to apply SAS. So the missing information is \(KJ\cong KT\).

Part (d)

Answer:

Step1: Recall ASA Congruence

ASA congruence requires two angles and the included side. In \(\triangle DEF\) and \(\triangle DGF\), we know that \(DF\) is common (so \(DF\cong DF\)) and \(\angle EDF=\angle GDF\) (marked angle). For ASA, we need the other pair of angles (the angles at \(E\) and \(G\)) to be congruent, i.e., \(\angle DEF\cong\angle DGF\) (or the angles at \(E\) and \(G\)) and also the included side. Wait, more precisely, in \(\triangle DEF\) and \(\triangle DGF\), we have \(DF\) as a common side, \(\angle EDF=\angle GDF\) (given angle). For ASA, we need \(\angle DFE\cong\angle DFG\) (the angles at \(F\))? No, wait, the triangles are \(\triangle DEF\) and \(\triangle DGF\). Let's label the angles: at \(D\), \(\angle EDF=\angle GDF\); side \(DF\) is common. For ASA, we need \(\angle DEF\cong\angle DGF\) (the angles at \(E\) and \(G\)) and the included side \(DF\). Wait, actually, the correct approach: ASA needs two angles and the included side. So we have \(\angle EDF=\angle GDF\) (angle), \(DF\) (side), and we need \(\angle DFE=\angle DFG\) (angle) or \(\angle DEF=\angle DGF\) (angle). But from the diagram, the quadrilateral \(DEFG\) is a kite - like figure? Wait, no, the triangles are \(\triangle DEF\) and \(\triangle DGF\). So to apply ASA, we need \(\angle DEF\cong\angle DGF\) (the angles at \(E\) and \(G\)) and the included side \(DF\) and angle \(\angle EDF=\angle GDF\). So the missing information is that \(\angle DEF\cong\angle DGF\) (or \(\angle DFE\cong\angle DFG\)) to apply ASA.

Final Answers:

(a)

To show \(\triangle FJG\cong\triangle HJG\) by SAS, we need \(FJ\cong HJ\) (or \(FJ = HJ\)).

(b)

To show \(\triangle VWX\cong\triangle ZYX\) by ASA, we need \(\angle VWX\cong\angle ZYX\) (or \(\angle V\cong\angle Z\)).

(c)

To show \(\triangle KJL\cong\triangle KTS\) by SAS, we need \(KJ\cong KT\).

(d)

To show \(\triangle DEF\cong\triangle DGF\) by ASA, we need \(\angle DEF\cong\angle DGF\) (or \(\angle DFE\cong\angle DFG\)).

(Note: The above is a detailed step - by - step for each sub - part following the Step - by - Step Format as the problem involves geometric deductions related to triangle congruence, which falls under the Mathematics discipline, subfield Geometry.)