Hello, I need help with this geometry question.
1 answer:
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The given triangles ΔDOA and ΔABC are right triangles that have a
common vertex at point <em>A</em>.
- <u>ΔDOA is similar to ΔABC, and </u><u>∠AEC</u><u> is equal to </u><u>∠ABC,</u><u> therefore, ∠AEC = ∠ODA</u>
Reasons:
The given parameters are;
The diameter of the circle with center <em>O</em> = AB
DO ⊥ AB (DO is perpendicular to AB)
Required:
Prove that ∠AEC = ∠ODA
A two column proof is presented as follows;
Statement Reason
1. AB is the diameter of circle <em> </em> 1. Given
2. DO is perpendicular to AB <em> </em> 2. Given
3. ∠DOA = 90° <em> </em> 3. Definition of DO ⊥ AB
4. ∠BCA = 90° <em> </em> 4. Thales theorem
5. ∠BCA ≅ ∠BCA <em> </em> 5. Reflexive property
6. ΔDOA ~ ΔABC <em> </em> 6. AA similarity postulate
7. ∠ABC ≅ ∠ODA <em> </em> 7. CASTC
8. ∠ABC = ∠ODA <em> </em> 8. Definition of congruency
9. ∠AEC ≅ ∠ABC <em> </em> 9. Angles in the same segment
10. ∠AEC = ∠ABC <em> </em> 10. Definition of congruency
11. ∠AEC = ∠ODA <em> </em> 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: