Difference between revisions of "1993 AHSME Problems/Problem 2"

(Created page with "== Problem == <asy> draw((-5,0)--(5,0)--(2,14)--cycle,black+linewidth(.75)); draw((-2.25,5.5)--(4,14/3),black+linewidth(.75)); MP("A",(-5,0),S);MP("C",(5,0),S);MP("B",(2,14),N);M...")
 
(Problem)
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<math>\text{(A)} 50^\circ\quad
+
<math>\text{(A) } 50^\circ\quad
\text{(B)} 55^\circ\quad
+
\text{(B) } 55^\circ\quad
\text{(C)} 60^\circ\quad
+
\text{(C) } 60^\circ\quad
\text{(D)} 65^\circ\quad
+
\text{(D) } 65^\circ\quad
\text{(E)} 70^\circ</math>
+
\text{(E) } 70^\circ</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 00:55, 26 September 2014

Problem

[asy] draw((-5,0)--(5,0)--(2,14)--cycle,black+linewidth(.75)); draw((-2.25,5.5)--(4,14/3),black+linewidth(.75)); MP("A",(-5,0),S);MP("C",(5,0),S);MP("B",(2,14),N);MP("E",(4,14/3),E);MP("D",(-2.25,5.5),W); MP("55^\circ",(-4.5,0),NE);MP("75^\circ",(5,0),NW);  [/asy]

In $\triangle ABC$, $\angle A=55^\circ$, $\angle C=75^\circ, D$ is on side $\overline{AB}$ and $E$ is on side $\overline{BC}$. If $DB=BE$, then $\angle{BED} =$


$\text{(A) } 50^\circ\quad \text{(B) } 55^\circ\quad \text{(C) } 60^\circ\quad \text{(D) } 65^\circ\quad \text{(E) } 70^\circ$

Solution

$\fbox{D}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AHSME Problems and Solutions

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