Question

which congruence theorems can be used to prove δabr ≅ δacr? select three options. image of triangles abr and acr with right angles at b and c, ab and ac marked congruent, br and cr marked congruent options: hl, sas, sss, asa, aas

Explanation:

Step1: Analyze HL (Hypotenuse - Leg)

In right - triangles \( \triangle ABR\) and \( \triangle ACR\), \(AR = AR\) (common hypotenuse) and \(AB = AC\) (given). So, by HL (Hypotenuse - Leg) congruence theorem, \( \triangle ABR\cong\triangle ACR\).

Step2: Analyze SAS (Side - Angle - Side)

We know that \(AB = AC\), \(\angle ABR=\angle ACR = 90^{\circ}\), and \(BR = CR\) (since \(R\) is equidistant from \(B\) and \(C\) as per the right - triangle and given equal segments). But also, we can note that \(AR\) is common. In terms of angles, \(\angle BAR=\angle CAR\) (if we consider the non - right angles). If we assume \(AB = AC\), \(AR = AR\) and \(\angle ABR=\angle ACR\) (right angles), we can also use the fact that for non - right angles, if we consider the triangles \( \triangle ABR\) and \( \triangle ACR\), \(AB = AC\), \( \angle ABR=\angle ACR\) (right angles) and \(BR = CR\) (by Pythagoras \(BR=\sqrt{AR^{2}-AB^{2}}\) and \(CR=\sqrt{AR^{2}-AC^{2}}\), since \(AB = AC\), \(BR = CR\)). But more simply, for SAS: \(AB = AC\), \( \angle ABR=\angle ACR\) (right angles) and \(BR = CR\) (derived or given in the geometric set - up).

Step3: Analyze AAS (Angle - Angle - Side)

We have two right angles (\(\angle ABR=\angle ACR = 90^{\circ}\)), a pair of equal non - right angles (say \(\angle BAR=\angle CAR\)) and a common side \(AR\). So, by AAS (Angle - Angle - Side) congruence theorem, \( \triangle ABR\cong\triangle ACR\).

Answer:

HL, SAS, AAS