Question

In figure 5.16, point X is mid-point of side BC, seg XZ || side AB and seg YZ || seg AX.
Prove that YC = $\frac{1}{4}$BC

Answer 1

$$\begin{gathered} \mathrm{I}\mathrm{n}∆\mathrm{ABC}\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{midpoint}\mathrm{of}\mathrm{BC} \\ \therefore \mathrm{XC}=\frac{1}{2}\mathrm{BC}...\left(1\right) \\ \mathrm{seg}\mathrm{XZ}\parallel \mathrm{side}\mathrm{AB} \\ \mathrm{By}\mathrm{converse}\mathrm{of}\mathrm{midpoint}\mathrm{theorem} \\ \mathrm{Z}\mathrm{is}\mathrm{the}\mathrm{midpoint}\mathrm{of}\mathrm{side}\mathrm{AC} \\ \mathrm{Now},\mathrm{consider}∆\mathrm{AXC}. \\ \mathrm{Z}\mathrm{is}\mathrm{the}\mathrm{midpoint}\mathrm{of}\mathrm{AC}\mathrm{and}\mathrm{seg}\mathrm{ZY}\parallel \mathrm{seg}\mathrm{AX} \\ \mathrm{By}\mathrm{converse}\mathrm{of}\mathrm{midpoint}\mathrm{theorem} \\ \mathrm{So},\mathrm{Y}\mathrm{is}\mathrm{the}\mathrm{midpoint}. \\ \therefore \mathrm{YC}=\frac{1}{2}\mathrm{XC} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{XC}\mathrm{from}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}: \\ \mathrm{YC}=\frac{1}{2}\left(\frac{1}{2}\mathrm{BC}\right)=\frac{1}{4}\mathrm{BC} \end{gathered}$$