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.
