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Answer:
(c) CPCTC
Step-by-step explanation:
The proof might go something like this:
- EH ≅ KH, FH ≅ JH . . . . . given
- ∠EHF ≅ ∠KHJ . . . . . vertical angles are congruent
- ∆EHF ≅ ∆KHJ . . . . . SAS congruence theorem
- EF ≅ KJ . . . . . CPCTC
Note that JK is an alternate name for segment KJ. We used KJ in the proof because vertices K and J correspond directly to vertices E and F in the congruent triangles.
