Answer by Hosam Hajeer for Angles created by triangle median
From the law of sines$ \dfrac{ AD }{\sin 20^\circ } = \dfrac{ DC }{\sin y } $and$ \dfrac{ AD}{\sin 40^\circ } = \dfrac{ BD }{\sin x } $Since $ DC = BD $ , then dividing out the two equations gives$...
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Angles created by triangle median
Given that $AD$ bisects the line $BC$ into equals parts $BD$ and $DC$, does that tell us anything about the angles $x$ and $y$?We also know $\angle B$ and $\angle C$. Is all this enough info to find...
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