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Difference between revisions of "1996 AHSME Problems/Problem 21"
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<math>2\angle C + \angle D + \angle A = 270^\circ</math>
 
<math>2\angle C + \angle D + \angle A = 270^\circ</math>
  
−
From isosceles <math>/triangle ABD</math>, we get <math>\angle A = \angle D</math>, so:
+
From isosceles <math>\triangle ABD</math>, we get <math>\angle A = \angle D</math>, so:
  
 
<math>2\angle C + 2\angle D = 270^\circ</math>
 
<math>2\angle C + 2\angle D = 270^\circ</math>
  
−
$\angle C + \angle D = 135^\circ
+
<math>\angle C + \angle D = 135^\circ</math>, which is answer <math>\boxed{D}</math>
  
 
==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 14:02, 20 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

Redraw the figure as a concave pentagon $ADECB$:

[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--E--C--B--A); markscalefactor=0.005; draw(rightanglemark(D, E, C)); 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]

The angles of the pentagon will still sum to $180^\circ \cdot 3 = 540^\circ$, regardless of whether the pentagon is concave or not. As a quick proof, note that the nine angles of three original triangles $\triangle AEB$, $\triangle CBE$, and $\triangle DEA$ all make up the angles of the pentagon without overlap.

Since reflex $\angle E = 270^\circ$, we have:

$\angle C + \angle B + \angle A + \angle D = 540^\circ - 270^\circ = 270^\circ$.

From isosceles $\triangle ABC$, we get $\angle B = \angle C$, so:

$2\angle C + \angle D + \angle A = 270^\circ$

From isosceles $\triangle ABD$, we get $\angle A = \angle D$, so:

$2\angle C + 2\angle D = 270^\circ$

$\angle C + \angle D = 135^\circ$, which is answer $\boxed{D}$

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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
All AHSME Problems and Solutions
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