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Difference between revisions of "1977 AHSME Problems/Problem 4"

(Created page with "==Solution== Solution by e_power_pi_times_i Because <math>\measuredangle A=80^\circ</math>, <math>\measuredangle B=\measuredangle C=50^\circ</math>. Then, because triangles <...")
 
(→‎Solution)
 
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== Problem 4 ==
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<asy>
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size(130);
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pair A = (2, 2.4), C = (0, 0), B = (4.3, 0),
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E = 0.7*A, F = 0.57*A + 0.43*B, D = (2.4, 0);
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draw(A--B--C--cycle);
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draw(E--D--F);
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label("$A$", A, N);
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label("$B$", B, E);
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label("$C$", C, W);
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label("$D$", D, S);
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label("$E$", E, NW);
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label("$F$", F, NE);
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//Credit to MSTang for the diagram
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</asy>
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In triangle <math>ABC, AB=AC</math> and <math>\measuredangle A=80^\circ</math>. If points <math>D, E</math>, and <math>F</math> lie on sides <math>BC, AC</math> and <math>AB</math>, respectively, and <math>CE=CD</math> and <math>BF=BD</math>, then <math>\measuredangle EDF</math> equals
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<math>\textbf{(A) }30^\circ\qquad
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\textbf{(B) }40^\circ\qquad
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\textbf{(C) }50^\circ\qquad
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\textbf{(D) }65^\circ\qquad
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\textbf{(E) }\text{none of these}  </math>
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==Solution==
 
==Solution==
 
Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
 
Because <math>\measuredangle A=80^\circ</math>, <math>\measuredangle B=\measuredangle C=50^\circ</math>. Then, because triangles <math>CDE</math> and <math>BDF</math> are isosceles, <math>\measuredangle CDE=\measuredangle BDF=65^\circ</math>. <math>\measuredangle EDF=180^\circ - 2(65^\circ)=\textbf{(C) }50^\circ</math>
 
Because <math>\measuredangle A=80^\circ</math>, <math>\measuredangle B=\measuredangle C=50^\circ</math>. Then, because triangles <math>CDE</math> and <math>BDF</math> are isosceles, <math>\measuredangle CDE=\measuredangle BDF=65^\circ</math>. <math>\measuredangle EDF=180^\circ - 2(65^\circ)=\textbf{(C) }50^\circ</math>

Latest revision as of 12:26, 21 November 2016

Problem 4

[asy] size(130); pair A = (2, 2.4), C = (0, 0), B = (4.3, 0), E = 0.7*A, F = 0.57*A + 0.43*B, D = (2.4, 0); draw(A--B--C--cycle); draw(E--D--F); label("$A$", A, N); label("$B$", B, E); label("$C$", C, W); label("$D$", D, S); label("$E$", E, NW); label("$F$", F, NE); //Credit to MSTang for the diagram [/asy]

In triangle $ABC, AB=AC$ and $\measuredangle A=80^\circ$. If points $D, E$, and $F$ lie on sides $BC, AC$ and $AB$, respectively, and $CE=CD$ and $BF=BD$, then $\measuredangle EDF$ equals

$\textbf{(A) }30^\circ\qquad \textbf{(B) }40^\circ\qquad \textbf{(C) }50^\circ\qquad \textbf{(D) }65^\circ\qquad \textbf{(E) }\text{none of these}$


Solution

Solution by e_power_pi_times_i

Because $\measuredangle A=80^\circ$, $\measuredangle B=\measuredangle C=50^\circ$. Then, because triangles $CDE$ and $BDF$ are isosceles, $\measuredangle CDE=\measuredangle BDF=65^\circ$. $\measuredangle EDF=180^\circ - 2(65^\circ)=\textbf{(C) }50^\circ$

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