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Question
a diagram is shown where (overleftrightarrow{fp}) intersects (overleftrightarrow{rt}), (overleftrightarrow{hd}), and (overleftrightarrow{zk}).
given: (rtparallel zk), (angle tmvcongangle dva)
prove: (hdparallel zk)
| statement | reason |
|---|---|
| 2. (angle tmvcongangle pak) | select |
| 3. (angle tmvcongangle dva) | 3. given |
| 4. (angle pakcongangle dva) | select |
| 5. (overleftrightarrow{hd}paralleloverleftrightarrow{zk}) | select |
Step1: Identify the property for \(\angle TMV\cong\angle PAK\)
Since \(RT\parallel ZK\), by the Corresponding Angles Theorem, \(\angle TMV\cong\angle PAK\) (when two parallel lines are cut by a transversal, corresponding angles are congruent).
Step2: Justify \(\angle PAK\cong\angle DVA\)
We know that \(\angle TMV\cong\angle PAK\) (from Step 1) and \(\angle TMV\cong\angle DVA\) (given). By the Transitive Property of Congruence (\(a = b\) and \(b = c\) implies \(a = c\) for congruent angles), \(\angle PAK\cong\angle DVA\).
Step3: Prove \(HD\parallel ZK\)
Since \(\angle PAK\cong\angle DVA\), by the Converse of the Corresponding Angles Theorem (if two lines are cut by a transversal such that corresponding angles are congruent, then the two lines are parallel), \(HD\parallel ZK\).
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- Corresponding Angles Theorem; 4. Transitive Property of Congruence; 5. Converse of the Corresponding Angles Theorem.