Question

Read the proof. Given: AB тИе DE Prove: тЦ│ACB ~ тЦ│DCE
We are given AB тИе DE. Because the lines are parallel and segment CB crosses both lines, we can consider segment CB a transversal of the parallel lines. Angles CED and CBA are corresponding angles of transversal CB and are therefore congruent, so тИаCED тЙЕ тИаCBA. We can state тИаC тЙЕ тИаC using the reflexive property. Therefore, тЦ│ACB ~ тЦ│DCE by the AA similarity theorem. SSS similarity theorem. AAS similarity theorem. ASA similarity theorem.

Answer 1

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