Look for a pair of vertical angles and prove that △ ABC ≅ △ EDC.
See solution.
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Since AE⊥ DE and EA⊥ AB, we have that m∠E = 90^(∘) and m∠A = 90^(∘). Therefore, ∠A ≅ ∠E. Also, AC ≅ EC because of the Midpoint Theorem.
Notice that ∠BCA and ∠DCE are vertical angles, and by the Vertical Angles Theorem we get ∠BCA ≅ ∠DCE.
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∠A ≅ ∠E & Angle
AC ≅ EC & Included Side
∠BCA ≅ ∠DCE & Angle
Finally, by the Angle-Side-Angle (ASA) Congruence Postulate, we obtain that △ ABC ≅ △ EDC which, by definition of congruent polygons, tells us that CD ≅ CB.
Paragraph Proof
Given: & AE⊥ DE, EA⊥ AB
& Cis the midpoint ofAE
Prove: & CD≅ CB
Proof: Since AE⊥ DE and EA⊥ AB, we have that m∠E = 90^(∘) and m∠A = 90^(∘). Therefore, ∠A ≅ ∠E. Also, AC ≅ EC because of the Midpoint Theorem. Additionally, ∠BCA and ∠DCE are vertical angles, and by the Vertical Angles Theorem, we get ∠BCA ≅ ∠DCE.
