Let $ABC$ be a triangle where $AB = 10$ and $AC = 12$. The bisectors of ${\angle}B$ and ${\angle}C$ intersect $AC$ and $AB$ at points $D$ and $E$ respectively. We draw perpendiculars $AM$ and $AN$ from vertex $A$ on line segments $CE$ and $BD$ respectively. Knowing that the length of $MN$ is $4$, how would one find the length of $BC$?
Let $\overrightarrow {AN} \cap \overline {BC} = \{F\}, \; \overrightarrow {AM} \cap \overline {BC} = \{G\}$.
$$\text{In} \ \Delta \text{BNA}, \Delta \text{BNF}: \\ \ m(\angle ABN) = m(\angle FBN), \\ m(\angle ANB) = m(\angle FNB), \\ \overline {BN} \text{ is a common side.} \\ \therefore \Delta \text{BNA} \equiv\Delta \text{BNF} \qquad \text{(1)}$$
$$\text{In} \ \Delta \text{CMA}, \Delta \text{CMG}: \\ m(\angle ACM) = m(\angle GCM), \\ m(\angle AMC) = m(\angle GMC), \\ \overline {CM} \text{ is a common side.} \\ \therefore \Delta \text{CMA} \equiv\Delta \text{CMG} \qquad \text{(2)}$$ From $(1), (2)$ we deduce that: $$AB=BF=10 \qquad \text{(3)} \\ AC=CG=12 \qquad \text{(4)}$$ $$\text{In} \Delta \text{AMN}, \Delta \text{AGF}: \\ AN=NF, \\ AM=MG, \\ \angle A \text{ is common.} \\ \therefore \Delta \text{AMN} \sim \Delta \text{AGF}$$
$$AM = MG, \\ AN = NF, \\ MN =4 \\ \therefore FG = 2MN = 8 \qquad \text{(5)}$$ From $(3), (4)$ and $(5)$ we conclude that: $\fbox{BC = 14}$