Question

1. if ab ⊥ bc, ed ⊥ dc and c is the midpoint of bd, is δcba ≅ δcde? explain your reasoning.

Explanation:

Step1: Identify right angles

Since \( AB \perp BC \) and \( ED \perp DC \), we have \( \angle ABC = 90^\circ \) and \( \angle EDC = 90^\circ \), so \( \angle ABC \cong \angle EDC \).

Step2: Identify midpoint segment

Given \( C \) is the midpoint of \( BD \), then \( BC = DC \).

Step3: Identify vertical angles

\( \angle ACB \) and \( \angle ECD \) are vertical angles, so \( \angle ACB \cong \angle ECD \).

Step4: Apply ASA congruence

By the Angle - Side - Angle (ASA) congruence criterion, in \( \triangle CBA \) and \( \triangle CDE \):

• \( \angle ABC \cong \angle EDC \) (right angles)
• \( BC = DC \) (midpoint definition)
• \( \angle ACB \cong \angle ECD \) (vertical angles)

So, \( \triangle CBA \cong \triangle CDE \) by ASA.

Answer:

Yes, \( \triangle CBA \cong \triangle CDE \) by the ASA (Angle - Side - Angle) congruence criterion. The right angles \( \angle ABC \) and \( \angle EDC \) are congruent, \( BC = DC \) (since \( C \) is the midpoint of \( BD \)), and the vertical angles \( \angle ACB \) and \( \angle ECD \) are congruent, satisfying the ASA conditions for triangle congruence.