Question

find the m∠bac, if m∠dec = 45° and m∠edc = 65°. options: 50°, 60°, 70°, 80°. diagram with ( overline{ab} parallel overline{cd} ), ( overline{bc} parallel overline{de} ), triangles ( abc ) and ( cde ).

Explanation:

Step1: Find ∠DCE in △DEC

In triangle \( DEC \), the sum of interior angles is \( 180^\circ \). So, \( m\angle DCE = 180^\circ - m\angle DEC - m\angle EDC \).
Substitute \( m\angle DEC = 45^\circ \) and \( m\angle EDC = 65^\circ \):
\( m\angle DCE = 180^\circ - 45^\circ - 65^\circ = 70^\circ \).

Step2: Use parallel lines to find ∠BAC

Since \( \overline{BC} \parallel \overline{DE} \), \( \angle BCA = \angle DCE \) (alternate interior angles). So \( m\angle BCA = 70^\circ \).
Also, \( \overline{AB} \parallel \overline{CD} \), so \( \angle BAC = \angle DCA \)? Wait, no, wait. Wait, actually, let's re - check. Wait, in triangle \( ABC \) and the parallel lines: Wait, actually, since \( AB\parallel CD \) and \( BC\parallel DE \), the triangles \( ABC \) and \( CDE \) are similar? Wait, no, more simply, since \( BC\parallel DE \), \( \angle BCA=\angle DCE \) (alternate interior angles). Then, since \( AB\parallel CD \), \( \angle BAC=\angle DCA \)? Wait, no, maybe a better way: In triangle \( ABC \), we can find \( \angle BAC \) by using the fact that \( AB\parallel CD \), so \( \angle BAC = \angle DCA \)? Wait, no, let's correct. Wait, first, in \( \triangle DEC \), we found \( \angle DCE = 70^\circ \). Then, since \( BC\parallel DE \), \( \angle BCA=\angle DCE = 70^\circ \)? No, wait, no. Wait, \( BC\parallel DE \), so \( \angle BCD=\angle CDE = 65^\circ \)? Wait, I think I made a mistake earlier. Let's start over.

Wait, in \( \triangle DEC \), angles sum to \( 180^\circ \). So \( \angle DCE=180 - 45 - 65 = 70^\circ \). Now, since \( BC\parallel DE \), \( \angle BCA=\angle DEC = 45^\circ \)? No, that's not right. Wait, the correct approach: Since \( BC\parallel DE \), \( \angle BCD=\angle CDE = 65^\circ \) (alternate interior angles). And since \( AB\parallel CD \), \( \angle BAC=\angle DCA \). Wait, maybe the triangles are congruent or similar. Wait, another way: The sum of angles in a triangle is \( 180^\circ \). Let's look at \( \triangle ABC \) and \( \triangle CDE \). Since \( AB\parallel CD \) and \( BC\parallel DE \), \( \angle ABC=\angle BCD \) and \( \angle BCD=\angle CDE \), so \( \angle ABC=\angle CDE = 65^\circ \), and \( \angle BCA=\angle DEC = 45^\circ \). Then in \( \triangle ABC \), \( \angle BAC=180 - \angle ABC-\angle BCA=180 - 65 - 45 = 70^\circ \). Wait, that makes sense. So first, find \( \angle DCE \) is wrong. Let's do it correctly:

In \( \triangle DEC \), \( \angle DEC = 45^\circ \), \( \angle EDC = 65^\circ \), so \( \angle DCE=180-(45 + 65)=70^\circ \). Now, since \( BC\parallel DE \), \( \angle BCA=\angle DEC = 45^\circ \)? No, that's incorrect. Wait, the correct alternate interior angles: If \( BC\parallel DE \), then \( \angle BCD=\angle CDE = 65^\circ \) (because \( BC \) and \( DE \) are parallel, and \( CD \) is a transversal). Then, since \( AB\parallel CD \), \( \angle BAC=\angle DCA \). Wait, \( \angle DCA \) is equal to \( \angle DCE \)? No, \( \angle DCA \) and \( \angle DCE \) are adjacent angles? Wait, the line \( AE \) is straight, so \( \angle ACB+\angle BCD+\angle DCE = 180^\circ \)? No, \( A - C - E \) is a straight line, so \( \angle ACB+\angle BCD+\angle DCE = 180^\circ \)? No, \( A \), \( C \), \( E \) are colinear, so \( \angle ACB+\angle BCD+\angle DCE = 180^\circ \)? No, actually, \( \angle ACB \) and \( \angle BCD \) and \( \angle DCE \) are on the straight line \( AE \), so their sum is \( 180^\circ \). But maybe a better approach: Since \( AB\parallel CD \), \( \angle BAC=\angle DCA \) (alternate interior angles). And since \( BC\parallel DE \), \( \angle BCA=\angle DEC = 45^\circ \) (alternate interior angles). Wait, no, \( BC\parallel DE \), so \( \angle BCA=\angle DEC \) (if \( AC \) is the transversal). Then in \( \triangle ABC \), \( \angle BAC = 180^\circ-\angle ABC - \angle BCA \). But \( \angle ABC=\angle BCD \) (because \( AB\parallel CD \), alternate interior angles), and \( \angle BCD=\angle CDE = 65^\circ \) (because \( BC\parallel DE \), alternate interior angles). So \( \angle ABC = 65^\circ \), \( \angle BCA=\angle DEC = 45^\circ \). Then \( \angle BAC=180 - 65 - 45=70^\circ \).

Answer:

\( 70^\circ \) (corresponding to the option with \( 70^\circ \))