The given triangles ΔDOA and ΔABC are right triangles that have a
common vertex at point A.
- ΔDOA is similar to ΔABC, and ∠AEC is equal to ∠ABC, therefore, ∠AEC = ∠ODA
Reasons:
The given parameters are;
The diameter of the circle with center O = AB
DO ⊥ AB (DO is perpendicular to AB)
Required:
Prove that ∠AEC = ∠ODA
A two column proof is presented as follows;
Statement [tex]{}[/tex] Reason
1. AB is the diameter of circle [tex]{}[/tex] 1. Given
2. DO is perpendicular to AB [tex]{}[/tex] 2. Given
3. ∠DOA = 90° [tex]{}[/tex] 3. Definition of DO ⊥ AB
4. ∠BCA = 90° [tex]{}[/tex] 4. Thales theorem
5. ∠BCA ≅ ∠BCA [tex]{}[/tex] 5. Reflexive property
6. ΔDOA ~ ΔABC [tex]{}[/tex] 6. AA similarity postulate
7. ∠ABC ≅ ∠ODA [tex]{}[/tex] 7. CASTC
8. ∠ABC = ∠ODA [tex]{}[/tex] 8. Definition of congruency
9. ∠AEC ≅ ∠ABC [tex]{}[/tex] 9. Angles in the same segment
10. ∠AEC = ∠ABC [tex]{}[/tex] 10. Definition of congruency
11. ∠AEC = ∠ODA [tex]{}[/tex] 11. Transitive property of equality
In statement 6, ΔDOA is similar to ΔABC by Angle-Angle, AA, similarity
postulate, therefore, the three angles of ΔDOA are congruent to the three
angles of ΔABC.
Therefore ∠ABC ≅ ∠ODA by Corresponding Angles of Similar Triangles
are Congruent, CASTC.
Learn more about circle theorem here:
https://brainly.com/question/16879446
