Difference between revisions of "2014 AMC 8 Problems/Problem 9"
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| + | ==Problem== | ||
In <math>\bigtriangleup ABC</math>, <math>D</math> is a point on side <math>\overline{AC}</math> such that <math>BD=DC</math> and <math>\angle BCD</math> measures <math>70^\circ</math>. What is the degree measure of <math>\angle ADB</math>? | In <math>\bigtriangleup ABC</math>, <math>D</math> is a point on side <math>\overline{AC}</math> such that <math>BD=DC</math> and <math>\angle BCD</math> measures <math>70^\circ</math>. What is the degree measure of <math>\angle ADB</math>? | ||
| − | <asy>size(300);defaultpen(linewidth(0.8));pair A=(-1,0),C=(1,0),B=dir(40),D=origin;draw(A--B--C--A);draw(D--B);dot("$A$" | + | <asy> |
| + | size(300); | ||
| + | defaultpen(linewidth(0.8)); | ||
| + | pair A=(-1,0),C=(1,0),B=dir(40),D=origin; | ||
| + | draw(A--B--C--A); | ||
| + | draw(D--B); | ||
| + | dot("$A$", A, SW); | ||
| + | dot("$B$", B, NE); | ||
| + | dot("$C$", C, SE); | ||
| + | dot("$D$", D, S); | ||
| + | label("$70^\circ$",C,2*dir(180-35));</asy> | ||
<math>\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150</math> | <math>\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150</math> | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/jLnqUOe0HPE | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=HP-lBKohxhE ~David | ||
| + | |||
| + | https://youtu.be/j5KrHM81HZ8 ~savannahsolver | ||
| + | |||
| + | ==Video Solution == | ||
| + | https://youtu.be/abSgjn4Qs34?t=3140 | ||
| + | |||
| + | ==Solution== | ||
| + | Using angle chasing is a good way to solve this problem. <math>BD = DC</math>, so <math>\angle DBC = \angle DCB = 70</math>, because it is an isosceles triangle. Then <math>\angle CDB = 180-(70+70) = 40</math>. Since <math>\angle ADB</math> and <math>\angle BDC</math> are supplementary, <math>\angle ADB = 180 - 40 = \boxed{\textbf{(D)}~140}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2014|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:32, 16 June 2024
Contents
Problem
In 
, 
is a point on side 
such that 
and 
measures 
. What is the degree measure of 
?
![[asy] size(300); defaultpen(linewidth(0.8)); pair A=(-1,0),C=(1,0),B=dir(40),D=origin; draw(A--B--C--A); draw(D--B); dot("$A$", A, SW); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, S); label("$70^\circ$",C,2*dir(180-35));[/asy]](http://latex.artofproblemsolving.com/9/4/7/94770df9a0e1d5a71f602bfb3b45022dcca9dec4.png)
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=HP-lBKohxhE ~David
https://youtu.be/j5KrHM81HZ8 ~savannahsolver
Video Solution
https://youtu.be/abSgjn4Qs34?t=3140
Solution
Using angle chasing is a good way to solve this problem. 
, so 
, because it is an isosceles triangle. Then 
. Since and 
are supplementary, 
.
See Also
| 2014 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
