Difference between revisions of "1965 AHSME Problems/Problem 25"
(created solution page) |
m (fixed typo) |
||
| Line 4: | Line 4: | ||
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> | + | \textbf{(E) }\ \text{one pair of equal angles} </math> |
== Solution == | == Solution == | ||
Revision as of 19:35, 18 July 2024
Problem
Let 
be a quadrilateral with 
extended to 
so that 
.
Lines 
and 
are drawn to form 
. For this angle to be a right angle it is necessary that quadrilateral have:
Solution
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
