In triangle $ABC$, $\angle BAC = 94^{\circ},\ \angle ACB = 39^{\circ}$. Prove that \[ BC^2 = AC^2 + AC\cdot AB\].
Problem
Source: Indian Postal Coaching 2012 Set 1 #5
Tags: geometry, trigonometry, circumcircle, geometry unsolved
Source: Indian Postal Coaching 2012 Set 1 #5
Tags: geometry, trigonometry, circumcircle, geometry unsolved
In triangle $ABC$, $\angle BAC = 94^{\circ},\ \angle ACB = 39^{\circ}$. Prove that \[ BC^2 = AC^2 + AC\cdot AB\].