Answer:
\[\frac{AX}{AB}=\frac{1}{4}\] i.e., \[AX=1K,\text{ }AB=4K\] (K- constant) \[\therefore BX=AB-AX\] \[=4K-1K=3K\] Now, \[\frac{AX}{XB}=\frac{1K}{3K}=\frac{1}{3}\] And, \[\frac{AY}{YC}=\frac{2}{6}=\frac{1}{3}\] \[\frac{AX}{XB}=\frac{AY}{YC}\] \[\therefore XY\parallel BC\] (By converse of Thales? theorem)
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