Question

directions: read the description of each question carefully. be sure to observe any markings that may appear on the diagrams.

1. given the right triangles shown at the right. δfgc is isosceles. which of the following methods will prove δabc congruent to δdef?

choose:
○ sss ○ hl
○ aas ○ sas

1. given the triangles abc and def as shown at the right. in addition to the markings, af = cd. which of the following methods can not be used to prove the triangles congruent?

choose
○ sas ○ sss

Explanation:

Question 1

Step 1: Analyze given info

We have right triangles \( \triangle ABC \) and \( \triangle DEF \), and \( \triangle FGC \) is isosceles, so \( FG = CG \). Also, from the diagram, \( AB = DE \) (marked equal) and \( \angle B=\angle E = 90^\circ \).

Step 2: Determine congruence method

We know \( \angle B=\angle E = 90^\circ \), \( AB = DE \), and we can show \( BC = EF \) (since \( FG = CG \) and adding the other segments). Wait, actually, let's check the methods. The HL (Hypotenuse - Leg) theorem applies to right triangles. Let's see: if we can show hypotenuse and leg equal. But also, AAS: two angles and a non - included side. Wait, since \( \triangle FGC \) is isosceles, \( \angle GFC=\angle GCF \), so \( \angle ABC=\angle DEF = 90^\circ \), \( \angle BCA=\angle EFD \) (since \( \angle GFC=\angle GCF \), so their supplements are equal), and \( AB = DE \). So by AAS (Angle - Angle - Side), we can prove congruence. But wait, let's re - evaluate. Wait, the right angles, one pair of legs equal (\( AB = DE \)), and the angles at \( C \) and \( F \) related. Alternatively, since \( \triangle FGC \) is isosceles, \( FG = CG \), and if we consider the right triangles, \( AB = DE \) (leg), \( \angle B=\angle E = 90^\circ \), and \( \angle ACB=\angle DFE \) (because \( \angle GCF=\angle GFC \), so \( 90^\circ-\angle GCF = 90^\circ-\angle GFC \), i.e., \( \angle ACB=\angle DFE \)). So two angles (\( \angle B=\angle E \), \( \angle ACB=\angle DFE \)) and a side (\( AB = DE \)): AAS. But also, let's check HL. Wait, maybe the correct method is AAS? Wait, no, let's look again. Wait, the diagram has \( AB \) and \( DE \) marked equal (legs), \( \angle B \) and \( \angle E \) are right angles, and \( \angle ACB=\angle DFE \) (from isosceles \( \triangle FGC \)). So AAS: two angles (right angle and the other acute angle) and a side (the leg \( AB = DE \)). So the method is AAS.

Step 1: Analyze given info

We have \( \triangle ABC \) and \( \triangle DEF \), \( AF = CD \), so \( AF+FC=CD + FC\), which means \( AC = DF \). Also, \( \angle B=\angle E = 90^\circ \), and \( BC = EF \) (marked equal).

Step 2: Check congruence methods

• SAS: We have \( BC = EF \), \( \angle B=\angle E = 90^\circ \), and \( AB = DE \) (wait, no, \( AC = DF \) from \( AF = CD \)). Wait, \( BC = EF \), \( \angle B=\angle E \), and \( AB = DE \)? No, \( AF = CD\Rightarrow AC = DF \). Wait, \( \angle B=\angle E = 90^\circ \), \( BC = EF \), and \( AC = DF \)? No, \( AC \) and \( DF \) are hypotenuses. Wait, let's list the sides. If \( AF = CD \), then \( AC=AF + FC\) and \( DF=DC + CF\), so \( AC = DF \). We have \( \angle B=\angle E = 90^\circ \), \( BC = EF \) (marked), and \( AC = DF \). So by HL (for right triangles, hypotenuse \( AC = DF \) and leg \( BC = EF \)) or SAS (if we consider \( BC = EF \), \( \angle B=\angle E \), and \( AB = DE \)? Wait, no, \( AB \) and \( DE \) are legs. Wait, the options are SAS and SSS. Let's check SSS: For SSS, we need three sides equal. We know \( BC = EF \), \( AC = DF \) (from \( AF = CD \)), but do we know \( AB = DE \)? From the diagram, \( AB \) and \( DE \) are not marked as equal. So SSS cannot be used because we don't have all three sides equal. SAS: We have \( BC = EF \), \( \angle B=\angle E = 90^\circ \), and if we can show \( AB = DE \)? Wait, no, \( AC = DF \), \( BC = EF \), \( \angle B=\angle E \). Wait, maybe I made a mistake. Wait, the question is which method can NOT be used. Let's re - examine. If \( AF = CD \), then \( AC = DF \). We have \( \angle B=\angle E = 90^\circ \), \( BC = EF \). So by SAS: \( BC = EF \), \( \angle B=\angle E \), and \( AB = DE \)? No, \( AB \) and \( DE \) are not given. Wait, no, \( AC = DF \), \( BC = EF \), \( \angle B=\angle E \). Wait, SAS requires two sides and the included angle. The included angle for \( BC \) and \( AB \) is \( \angle B \), and for \( EF \) and \( DE \) is \( \angle E \). If \( BC = EF \), \( \angle B=\angle E \), and \( AB = DE \), but we don't know \( AB = DE \). Wait, no, from \( AF = CD \), \( AC = DF \), \( BC = EF \), \( \angle B=\angle E = 90^\circ \), so by HL (hypotenuse \( AC = DF \), leg \( BC = EF \)) they are congruent. Now, SSS: To use SSS, we need \( AB = DE \), \( BC = EF \), \( AC = DF \). We know \( BC = EF \) and \( AC = DF \), but we don't know \( AB = DE \) (it's not marked and not derived from \( AF = CD \)). So SSS cannot be used, while SAS can be used (if we consider \( BC = EF \), \( \angle B=\angle E \), and \( AB = DE \)? Wait, no, maybe I messed up. Wait, the correct answer is that SSS cannot be used, so the method that can NOT be used is SSS.

Answer:

AAS

Question 2