1996 AJHSME Problems/Problem 24

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 $\boxed{C}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1996 AJHSME Problems/Problem 24

Problem

Solution

See Also