Question

I need help with problems #1-3

Answer 1

Answer:

  see below

Step-by-step explanation:

In general, there are five ways to find if two triangles are congruent: SSS, SAS, ASA, AAS and HL.

SSA is another way that can be used if (and only if) the measure of the angle is 90° or is known to be the largest angle in the triangle. (That's another way of saying the first S in SSA must be the longest side.)

HL is a special case of SSA in which the angle is exactly 90°.

(1)

• ΔMXD ≅ ΔMYT . . . . by SAS congruence. Note the marks on the sides and the angle and the relationship between the angle and the sides—the angle is between the sides (SAS)
• ΔBNG (not congruent) ΔBXY . . . . this would require congruence by SSA, as the angle is not between the sides marked congruent. It appears that NG and XY are the longest sides. If that appearance can be taken at face value, then the conditions for use of SSA congruence are met and the triangles can be declared congruent. (Usually, the appearance of a figure cannot be assumed to mean anything. The only thing that matters is the markings.)
• ΔMPD (not congruent to anything) . . . . the marked angle on one side is not in the same relation to the marked side as it is on the other side of M.
• ΔNMT (not congruent) ΔTCN . . . . this would require congruence by SSA as the alternate interior angles ∠MNT and ∠CTN are not between side TN and the other sides marked congruent. The angles at M and C appear to be right angles, but are not marked as such, so we cannot conclude that HL congruence can be shown. The congruent angles are not opposite the longest side (TN), so the conditions for SSA congruence are not met.
• ΔITR ≅ ΔERT . . . . the relationship between the angles marked congruent and the common side (congruent to itself) TR is that of Angle, Angle, Side, so AAS congruence can be used.

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(2) See above for a list of congruence theorems. AAA can be used to show similarity, but not congruence.

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(3) See the discussion above. Only under certain circumstances can SSA be used as a "congruence shortcut." The 2nd and 4th examples above are partial illustration of this.

In the 4th example of problem (1), if the left and right parallel lines were skewed slightly more so that points C and M were closer together, it is easier to see that the line NC could be rotated about point N to a position where it would intersect line TC in a different place. Thus, even though the sides and angles are the same, there is no necessary congruence. (The attachment illustrates this.)