<emb and <ewb
B: angles and side lengths
To determine which overlapping triangles are congruent by the Angle-Side-Angle (ASA) postulate, you need to identify two angles and the included side of one triangle that correspond to two angles and the included side of another triangle. If both triangles share a side and have two pairs of equal angles, then they are congruent by ASA. For a specific example, if triangles ABC and DEF share side BC and have ∠A = ∠D and ∠B = ∠E, then triangles ABC and DEF are congruent by ASA.
To show that triangle ABC is congruent to triangle DEF by the Angle-Side-Angle (ASA) criterion, you need to establish that two angles and the included side of triangle ABC are congruent to the corresponding two angles and the included side of triangle DEF. Specifically, you would need to demonstrate that ∠A is congruent to ∠D, ∠B is congruent to ∠E, and the side AB is congruent to side DE. Once these conditions are satisfied, you can conclude that triangle ABC is congruent to triangle DEF by the ASA theorem.
If triangle DEC is congruent to triangle BEC by the principle of CPCTC (Corresponding Parts of Congruent Triangles are Congruent), then all corresponding sides and angles of the two triangles are equal. This means that side DE is equal to side BE, side EC is equal to side BC, and the angles ∠D and ∠B are congruent, as well as ∠E and ∠C. Thus, any corresponding part from one triangle can be stated to be congruent to its counterpart in the other triangle.
Yes. Read on for why: Take a parallelogram ABCD with midpoints E and F in the bases. So something like this (forgive the "drawing"): A E B __.__ /__.__/ C F D We know that parallelogram AEFC = EBDF, since they have the same base (F bisects CD, so CF = FD), height (haven't touched that), and angles (<ACF = <EFD because they're parallel - trust me that everything else matches). We also know that every parallelogram can be divided into two congruent triangles along their diagonal. So if two congruent parallelograms consistent of two congruent triangles each, then all four triangles are congruent. So your congruent triangles are ACF, AEF, EFD, and EBD. You can further reinforce this through ASA triangle congruency proofs (as I did at first), but this is a far more concise and equally valid answer.
B: angles and side lengths
To determine which overlapping triangles are congruent by the Angle-Side-Angle (ASA) postulate, you need to identify two angles and the included side of one triangle that correspond to two angles and the included side of another triangle. If both triangles share a side and have two pairs of equal angles, then they are congruent by ASA. For a specific example, if triangles ABC and DEF share side BC and have ∠A = ∠D and ∠B = ∠E, then triangles ABC and DEF are congruent by ASA.
Equilateral triangle or equiangular!
To show that triangle ABC is congruent to triangle DEF by the Angle-Side-Angle (ASA) criterion, you need to establish that two angles and the included side of triangle ABC are congruent to the corresponding two angles and the included side of triangle DEF. Specifically, you would need to demonstrate that ∠A is congruent to ∠D, ∠B is congruent to ∠E, and the side AB is congruent to side DE. Once these conditions are satisfied, you can conclude that triangle ABC is congruent to triangle DEF by the ASA theorem.
If triangle DEC is congruent to triangle BEC by the principle of CPCTC (Corresponding Parts of Congruent Triangles are Congruent), then all corresponding sides and angles of the two triangles are equal. This means that side DE is equal to side BE, side EC is equal to side BC, and the angles ∠D and ∠B are congruent, as well as ∠E and ∠C. Thus, any corresponding part from one triangle can be stated to be congruent to its counterpart in the other triangle.
Vertical angles are a pair of non-adjacent angles formed by the intersection of two lines. They are congruent, meaning they have the same measure. In other words, vertical angles are opposite each other when two lines intersect, and their measures are equal.
A. It is convex. D. Its sides are line segments. E. all of its sides are congruent. F. All of its angles are congruent.
Yes. Read on for why: Take a parallelogram ABCD with midpoints E and F in the bases. So something like this (forgive the "drawing"): A E B __.__ /__.__/ C F D We know that parallelogram AEFC = EBDF, since they have the same base (F bisects CD, so CF = FD), height (haven't touched that), and angles (<ACF = <EFD because they're parallel - trust me that everything else matches). We also know that every parallelogram can be divided into two congruent triangles along their diagonal. So if two congruent parallelograms consistent of two congruent triangles each, then all four triangles are congruent. So your congruent triangles are ACF, AEF, EFD, and EBD. You can further reinforce this through ASA triangle congruency proofs (as I did at first), but this is a far more concise and equally valid answer.
Oh, dude, angles in the letter E? Like, let me break it down for you. The letter E has three angles - two acute angles and one obtuse angle. So, next time you're doodling E's in your notebook, just remember, it's got some angles going on. Cool, right?
If triangles ABC and DEF are congruent (ABC ≅ DEF), then corresponding parts of the triangles are congruent by the principle of CPCTC (Corresponding Parts of Congruent Triangles are Congruent). This means that segments AB ≅ DE, BC ≅ EF, and AC ≅ DF, as well as angles ∠A ≅ ∠D, ∠B ≅ ∠E, and ∠C ≅ ∠F. All these congruences must be true if the triangles are indeed congruent.
If triangles ABC and DEF are congruent (ABC ≅ DEF), then by CPCTC (Corresponding Parts of Congruent Triangles are Congruent), the corresponding sides and angles are also congruent. This means that side (AB) is congruent to side (DE), side (BC) is congruent to side (EF), and side (AC) is congruent to side (DF). Additionally, angle (A) is congruent to angle (D), angle (B) is congruent to angle (E), and angle (C) is congruent to angle (F).
Then the two angles are said to be supplementary because they add up to 180 degrees.