Y by Adventure10, Mango247
Let
be an acute triangle. We draw
triangles
on the sides of
at the out sides such that:
![\[ \angle{B'AC}=\angle{C'BA}=\angle{A'BC}=30^{\circ} \ \ \ , \ \ \ \angle{B'CA}=\angle{C'AB}=\angle{A'CB}=60^{\circ} \]](//latex.artofproblemsolving.com/3/a/e/3aecf8834714a1615a8da632a5f6132863fb71f6.png)
If
is the midpoint of side
, prove that
is perpendicular to
.




![\[ \angle{B'AC}=\angle{C'BA}=\angle{A'BC}=30^{\circ} \ \ \ , \ \ \ \angle{B'CA}=\angle{C'AB}=\angle{A'CB}=60^{\circ} \]](http://latex.artofproblemsolving.com/3/a/e/3aecf8834714a1615a8da632a5f6132863fb71f6.png)
If




Stay ahead of learning milestones! Enroll in a class over the summer!
Something appears to not have loaded correctly.