Answer:
AC ≅ FD by CPCTC
Explanation:
You want to show that AC ≅ FD given triangles ABC and FED on segment BCDE with BD≅CE, AB≅FE, AB⊥BE, and EF⊥BE.
Proof
Step 2: BC +CD = CD +DE . . . . . segment addition theorem
Step 3: BC = DE . . . . . . . . . . subtraction property of equality
Step 4: ΔABC ≅ ΔFED . . . . . . . SAS congruence postulate
Step 5: AC ≅ FD . . . . . . . . . CPCTC
Discussion
Congruent segments BD and CE both contain segment CD, so that segment can be subtracted to leave BC ≅ DE. The right angles at B and E mean the angle between the congruent corresponding legs is the same:
AB≅FE, ∠B≅∠E, BD≅ED
This is the setup for claiming congruence by the SAS postulate. Since Corresponding Parts of Congruent Triangles are Congruent (CPCTC), hypotenuse AC is congruent to hypotenuse FD.
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Additional comment
It is useful to remember there are basically 4 ways to show triangles are congruent: AAS, ASA, SAS, SSS. When the triangles are right triangles, three of these are renamed, and one is added: AAS ⇒ HA, ASA ⇒ LA, SAS ⇒ LL, and the HL theorem is added.
In this problem, we made use of the LL theorem (SAS) , since one pair of legs was already marked, and the other pair was easy to show based on the given segment congruence.