Question

You have this information about ΔABC, ΔDEF, and ΔGHI:

Answer 1

ΔABC, ΔDEF, and ΔGHI

I think all triangles are congruent. 2 sides and 1 angle of each triangle is has the same measure. Making these triangles congruent in SAS theorem.

Answer 2

Answer: ΔABC, ΔDEF, and ΔGHI

Step-by-step explanation:

Given: In ΔABC, ΔDEF, and ΔGHI:

AB = DF               AB = GI

BC = HI                DE = HI

m∠B = m∠D = m∠I

In ΔABC and ΔGHI

AB = GI  [given]

BC = HI  [given]

m∠B =  m∠I  [given]

[ here m∠B and m∠I are the included angle of ΔABC and ΔGHI]

∴ ΔABC ≅ ΔGHI [by SAS congruence postulate]

In ΔABC and ΔDEF

AB = DF [given]

BC = DE [ Since BC = HI  and DE = HI so by transitive property BC = DE]

m∠B =  m∠D  [given]

[ here m∠B and m∠D are the included angle of ΔABC and ΔDEF]

∴ ΔABC ≅ ΔDEF [by SAS congruence postulate]

Now, since ΔABC ≅ ΔGHI and ΔABC ≅ ΔDEF

⇒  ΔGHI ≅ ΔDEF [transitive property]

Hence, all the given triangles ΔABC, ΔDEF, and ΔGHI are con gruent to each other.