Question

de is a mid - segment of isosceles triangle abc, and ac ≅ ab. which statements can be proven? select all that apply. a) ∠bca ≅ ∠cba b) ∠ade ≅ ∠abc c) ∠dae ≅ ∠ade d) bc = 2de e) ac = 2de

Explanation:

Step1: Recall isosceles - triangle property

In an isosceles triangle \(ABC\) with \(AC\cong AB\), the base - angles are equal. So, \(\angle BCA\cong\angle CBA\) (by the isosceles - triangle base - angle theorem).

Step2: Recall mid - segment theorem

Since \(DE\) is a mid - segment of \(\triangle ABC\), by the mid - segment theorem, \(DE\parallel BC\) and \(BC = 2DE\). Also, when \(DE\parallel BC\), corresponding angles are equal. So, \(\angle ADE\cong\angle ABC\) (corresponding angles for parallel lines \(DE\) and \(BC\) with transversal \(AB\)).

Answer:

A. \(\angle BCA\cong\angle CBA\)
B. \(\angle ADE\cong\angle ABC\)
D. \(BC = 2DE\)