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.
