Question

in the given figure ,o is the centr of the circle, ab is the diameter and do is perpendicular to ab. prove that angle AEC= ANGLE ODA​

Answer 1

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 ${}$                                                 Reason

1. AB is the diameter of circle     ${}$                 1. Given

2. DO is perpendicular to AB ${}$                     2. Given

3. ∠DOA = 90° ${}$                                             3. Definition of DO ⊥ AB

4. ∠BCA = 90°   ${}$                                             4. Thales theorem

5. ∠BCA ≅ ∠BCA ${}$                                         5. Reflexive property

6. ΔDOA ~ ΔABC   ${}$                                        6. AA similarity postulate

7. ∠ABC ≅ ∠ODA   ${}$                                       7. CASTC

8. ∠ABC = ∠ODA   ${}$                                        8. Definition of congruency

9. ∠AEC ≅ ∠ABC   ${}$                                        9. Angles in the same segment

10. ∠AEC = ∠ABC   ${}$                                       10. Definition of congruency

11. ∠AEC = ∠ODA   ${}$                                        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:

brainly.com/question/16879446