Difference between revisions of "1996 AHSME Problems/Problem 21"

(Created page with "==See also== {{AHSME box|year=1996|num-b=20|num-a=22}}")
 
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==Problem==
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Triangles <math>ABC</math> and <math>ABD</math> are isosceles with <math>AB=AC=BD</math>, and <math>BD</math> intersects <math>AC</math> at <math>E</math>. If <math>BD</math> is perpendicular to <math>AC</math>, then <math> \angle C+\angle D </math> is
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<asy>
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size(120);
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pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20);
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dot(A^^B^^C^^D^^E);
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draw(A--D--B--A--C--B);
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markscalefactor=0.005;
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draw(rightanglemark(A, E, B));
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dot(A^^B^^C^^D^^E);
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pair point=midpoint(A--M);
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label("$A$", A, dir(point--A));
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label("$B$", B, dir(point--B));
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label("$C$", C, dir(point--C));
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label("$D$", D, dir(point--D));
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label("$E$", E, dir(point--E));
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</asy>
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<math> \text{(A)}\ 115^\circ\qquad\text{(B)}\ 120^\circ\qquad\text{(C)}\ 130^\circ\qquad\text{(D)}\ 135^\circ\qquad\text{(E)}\ \text{not uniquely determined} </math>
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==Solution==
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==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=20|num-a=22}}
 
{{AHSME box|year=1996|num-b=20|num-a=22}}

Revision as of 21:02, 19 August 2011

Problem

Triangles $ABC$ and $ABD$ are isosceles with $AB=AC=BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC$, then $\angle C+\angle D$ is

[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); [/asy]

$\text{(A)}\ 115^\circ\qquad\text{(B)}\ 120^\circ\qquad\text{(C)}\ 130^\circ\qquad\text{(D)}\ 135^\circ\qquad\text{(E)}\ \text{not uniquely determined}$

Solution

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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