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Difference between revisions of "1996 AJHSME Problems/Problem 24"

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== Solution 2 ==
 
== Solution 2 ==
−
Contruct <math>\overline{BE}</math> through <math>textbf{D}</math> and intersects <math>\overline{AC}</math> at point <math>textbf{E}</math>
+
Contruct <math>\overline{BE}</math> through <math>D</math> and intersects <math>\overline{AC}</math> at point <math>E</math>
 +
 
 +
By Exterior Angle Theorem,
 +
 
 +
<math>\angle{ADE}</math> <math>=</math> <math>\angle{ABD} + \angle{BAD}</math>
 +
 
 +
Similarly,
 +
 
 +
<math>\angle{EDC} = \angle{DBC} + \angle{BCD}</math>
 +
 
 +
Thus,
 +
 
 +
<math>\angle{ADC} = \angle{ABC} + \angle{BAD} + \angle{BCD}</math>
 +
 
 +
Let <math>\angle{ADC} = x</math>
 +
 
 +
Because <math>\overline{AD}</math> and <math>\overline{CD}</math> are angle bisectors,
 +
 
 +
<math>180^\circ - x = \angle{BAD} + \angle{BCD}</math>
 +
 
 +
        <math>= x - 50^\circ</math>
 +
 
 +
<math>x = 115^\circ</math>
 +
 
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Thus, the answer is <math>\boxed{C}</math>
 +
 
 +
~ lovelearning999
  
 
==See Also==
 
==See Also==

Latest revision as of 07:41, 6 October 2024

Problem

The measure of angle $ABC$ is $50^\circ$, $\overline{AD}$ bisects angle $BAC$, and $\overline{DC}$ bisects angle $BCA$. The measure of angle $ADC$ is

[asy] pair A,B,C,D; A = (0,0); B = (9,10); C = (10,0); D = (6.66,3); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--cycle); draw(A--D--C); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,N); label("$50^\circ $",(9.4,8.8),SW); [/asy]

$\text{(A)}\ 90^\circ \qquad \text{(B)}\ 100^\circ \qquad \text{(C)}\ 115^\circ \qquad \text{(D)}\ 122.5^\circ \qquad \text{(E)}\ 125^\circ$

Solution

Let $\angle CAD = \angle BAD = x$, and let $\angle ACD = \angle BCD = y$

From $\triangle ABC$, we know that $50 + 2x + 2y = 180$, leading to $x + y = 65$.

From $\triangle ADC$, we know that $x + y + \angle D = 180$. Plugging in $x + y = 65$, we get $\angle D = 180 - 65 = 115$, which is answer $\boxed{C}$.

Solution 2

Contruct $\overline{BE}$ through $D$ and intersects $\overline{AC}$ at point $E$

By Exterior Angle Theorem,

$\angle{ADE}$ $=$ $\angle{ABD} + \angle{BAD}$

Similarly,

$\angle{EDC} = \angle{DBC} + \angle{BCD}$

Thus,

$\angle{ADC} = \angle{ABC} + \angle{BAD} + \angle{BCD}$

Let $\angle{ADC} = x$

Because $\overline{AD}$ and $\overline{CD}$ are angle bisectors,

$180^\circ - x = \angle{BAD} + \angle{BCD}$ 

       $= x - 50^\circ$

$x = 115^\circ$

Thus, the answer is $\boxed{C}$

~ lovelearning999

See Also

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
All AJHSME/AMC 8 Problems and Solutions

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