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

Latest revision as of 15:37, 14 July 2021

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

We first consider $\angle CBA$ . Because $\angle A = 55$ and $\angle C = 75$ , $\angle B = 180 - 55 - 75 = 50$ . Then, because $\triangle BED$ is isosceles, we have the equation $2 \angle BED + 50 = 180$ . Solving this equation gives us $\angle BED = 65 \rightarrow \fbox{\textbf{(D)}65}$

1993 AHSME (Problems • Answer Key • Resources)
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@@ Line 18: / Line 18: @@
 == Solution ==
-<math>\fbox{D}</math>
+We first consider <math>\angle CBA</math>. Because <math>\angle A = 55</math> and <math>\angle C = 75</math>, <math>\angle B = 180 - 55 - 75 = 50</math>. Then, because <math>\triangle BED</math> is isosceles, we have the equation <math>2 \angle BED + 50 = 180</math>. Solving this equation gives us <math>\angle BED = 65 \rightarrow \fbox{\textbf{(D)}65}</math>
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Difference between revisions of "1993 AHSME Problems/Problem 2"

Latest revision as of 15:37, 14 July 2021

Problem

Solution

See also