Difference between revisions of "1996 AJHSME Problems/Problem 24"
(→Solution 2) |
(→Solution 2) |
||
| Line 29: | Line 29: | ||
== Solution 2 == | == Solution 2 == | ||
Contruct <math>\overline{BE}</math> through <math>D</math> and intersects <math>\overline{AC}</math> at point <math>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, | By Exterior Angle Theorem, | ||
| − | <math>angle{ADE}</math> | + | |
| + | <math>\angle{ADE}</math> <math>=</math> <math>\angle{ABD} + \angle{DBC}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 21:18, 1 October 2024
Contents
Problem
The measure of angle 
is 
, 
bisects angle 
, and 
bisects angle 
. The measure of angle 
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]](http://latex.artofproblemsolving.com/b/8/c/b8c508ab617894b8e5319d685f1b7a1a01defa24.png)
Solution
Let 
, and let 
From 
, we know that 
, leading to 
.
From 
, we know that 
. Plugging in , we get 
, which is answer 
.
Solution 2
Contruct 
through 
and intersects 
at point 
By Exterior Angle Theorem,
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
