Question

state how the triangles are congruent using sss, sas, asa, aas, or hl. if they are not congruent, type not.
image of triangle with vertices g, f, h and i
type your answer...
10 fill in the blank 5 points
state how the triangles are congruent using sss, sas, asa, aas, or hl. if they are not congruent, type not.
image of another triangle
type your answer...
11 multiple choice 4 points
what else must you know to prove the triangles congruent by asa?
image of quadrilateral with vertices a, b, c, d and a triangle inside

Explanation:

First Triangle (Top)

Step1: Identify Common Side

Triangles \( \triangle GFH \) and \( \triangle IFH \) share side \( FH \), so \( FH = FH \) (reflexive property).

Step2: Identify Marked Sides and Angles

We see \( GF = IH \) (marked congruent) and \( \angle GFH = \angle IHF \) (marked congruent).

Step3: Apply SAS Criterion

We have two sides and the included angle: \( GF = IH \), \( \angle GFH = \angle IHF \), \( FH = FH \). So by SAS (Side - Angle - Side), the triangles are congruent.

Step1: Check Congruent Parts

We have two sides marked congruent, but the angle between them (for SAS) or the other corresponding parts don't match the congruence criteria. The given markings don't satisfy SSS, SAS, ASA, AAS, or HL.

Step2: Determine Congruence

Since the given information doesn't meet any triangle congruence criterion, the triangles are not congruent.

For ASA (Angle - Side - Angle), we need two angles and the included side to be congruent. In the diagram with quadrilateral \( ABCD \) (assuming the triangles are \( \triangle ABC \) and \( \triangle CDA \) or similar), we already have one pair of angles. To use ASA, we need to know that the included side between the two angles is congruent. For example, if we have \( \angle BAC \) and \( \angle DCA \), we need \( AC = CA \) (which is reflexive) and the other angle - side pair. But more precisely, we need to know that the side between the two known angles is congruent. If the triangles are \( \triangle AEC \) and \( \triangle BEC \) (assuming), we need the included side between the two angles to be congruent. In the context of the diagram (with \( A, B, C, D \)), to prove by ASA, we must know that \( AB \parallel CD \) (to get alternate interior angles) or that the included side (like \( AC \)) is congruent, but more accurately, we need the side between the two angles to be congruent. If the angles are \( \angle BAC \) and \( \angle DCA \), we need \( AC = CA \) (reflexive) and the other angle, but the key is the included side. So, for example, if we have two angles, we need the side between them to be congruent. So the answer would be that the included side between the two angles is congruent (e.g., \( AC = CA \) or the corresponding included side).

Answer:

SAS

Second Triangle (Middle)