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3 years ago

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

2 answers:

8 0

<span>ΔABC, ΔDEF, and ΔGHI
</span>
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.

zavuch27 [327]3 years ago

8 0

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.

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