The answer depends on what is already known about the two triangles.
"What else" implies there is already something that is congruent. But since you have not bothered to share that crucial bit of information, I cannot provide a more useful answer. no correct
To show that triangle ABC is congruent to triangle XYZ by the ASA (Angle-Side-Angle) criterion, we need to establish that two angles in triangle ABC are congruent to two angles in triangle XYZ, along with the side that is included between those angles being congruent. Specifically, if we have ∠A ≅ ∠X, ∠B ≅ ∠Y, and side AB ≅ XY, then the triangles can be concluded as congruent by ASA. Thus, we would need to confirm the congruence of these angles and the included side.
No, something could be similar to something else, but be bigger. To be congruent they need to be the exactly the same.
To show that triangle JKL is congruent to triangle MNO by the Angle-Angle-Side (AAS) theorem, you need to establish that two angles and the non-included side of triangle JKL are congruent to two angles and the corresponding non-included side of triangle MNO. Specifically, you would need to verify that one of the angles in triangle JKL is congruent to one of the angles in triangle MNO, and that the side opposite the angle in triangle JKL is congruent to the corresponding side in triangle MNO. This would complete the necessary conditions for AAS congruence.
__ - __ AC = XZ = is the similar sign
Angle "A" is congruent to Angle "D"
The answer depends on what is already known about the two triangles.
That depends on which sides have not been proven congruent yet.
For a start, you would need to know what efg and jkl are.
"What else" implies there is already something that is congruent. But since you have not bothered to share that crucial bit of information, I cannot provide a more useful answer.
Bc= qr
Su jL
bc yz
"What else" implies there is already something that is congruent. But since you have not bothered to share that crucial bit of information, I cannot provide a more useful answer.
Line segment BC is congruent to Line Segment YZ
We don't know what has already been proven congruent, sowe're in no position to be able to say what elseis required.