Prove that if triangle ABC and triangle DEF are triangles with equal defects and a pair of congruent sides, their associated Saccheri quadrilaterals are congruent. (HINT: Recall that the angle sum for a triangle is equal to the sum of the measures of the summit angles of its associated Saccheri quadrilateral and that the two summit angles of a Saccheri quadrilateral are congruent.)

According to theorem 7.5

Π ABB'A' ≅ Π DEE'D'  

therefore by transitivity of equivalence  it is proven that triangle ABC and triangle DEF are triangles with equal defects and a pair of congruent sides

Step-by-step explanation:

To prove that triangle ABC and triangle DEF are triangles with equal defects and a pair of congruent sides :

Assume:  б(Δ ABC ) =  б(Δ DEF ) and also AB ≅ DE

let Π ABB'A' and DEE'D' be taken as the saccheri quadrilaterals that corresponds to Δ ABC and Δ DEF  respectively

Following the Lemma above; б(Π ABB'A' ) = б( Π DEE'D' ) given that

AB = summit of ABB'A' and DE = summit of DEE'D'  also  AB ≅ DE

According to theorem 7.5

Π ABB'A' ≅ Π DEE'D'  

therefore by transitivity of equivalence  it is proven that triangle ABC and triangle DEF are triangles with equal defects and a pair of congruent sides

the answer is a

step-by-step explanation:

because your just changing the position of where the shape is