Art of Problem Solving

1965 AHSME Problems/Problem 25: Difference between revisions

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Lines <math>AC</math> and <math>CE</math> are drawn to form <math>\angle{ACE}</math>. For this angle to be a right angle it is necessary that quadrilateral <math>ABCD</math> have:  
Lines <math>AC</math> and <math>CE</math> are drawn to form <math>\angle{ACE}</math>. For this angle to be a right angle it is necessary that quadrilateral <math>ABCD</math> have:  


<math>\textbf{(A)}\ \text{all angles equal}  
<math>\textbf{(A)}\ \text{all angles equal} \qquad
\textbf{(B) }\ \text{all sides equal} \\
\textbf{(B) }\ \text{all sides equal} \\
\textbf{(C) }\ \text{two pairs of equal sides} \qquad
\textbf{(C) }\ \text{two pairs of equal sides} \qquad
\textbf{(D) }\ \text{one pair of equal sides} \\
\textbf{(D) }\ \text{one pair of equal sides} \\
\textbf{(E) }\ \text{one pair of equal angles} </math>   \qquad
\textbf{(E) }\ \text{one pair of equal angles} </math>


== Solution ==
== Solution ==

Revision as of 18:35, 18 July 2024

Problem

Let $ABCD$ be a quadrilateral with $AB$ extended to $E$ so that $\overline{AB} = \overline{BE}$. Lines $AC$ and $CE$ are drawn to form $\angle{ACE}$. For this angle to be a right angle it is necessary that quadrilateral $ABCD$ have:

$\textbf{(A)}\ \text{all angles equal} \qquad \textbf{(B) }\ \text{all sides equal} \\ \textbf{(C) }\ \text{two pairs of equal sides} \qquad \textbf{(D) }\ \text{one pair of equal sides} \\ \textbf{(E) }\ \text{one pair of equal angles}$

Solution

$\boxed{\textbf{(D)}}$


See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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All AHSME Problems and Solutions

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