In the given figure, points ^@ M ^@ and ^@ N ^@ divide the side ^@ AB ^@ of ^@\Delta ABC^@ into three equal parts. Line segments ^@MP^@ and ^@NQ^@ are both parallel to ^@BC^@ and meet ^@AC^@ at ^@P^@ and ^@Q^@ respectively. Prove that ^@P^@ and ^@Q^@ divide ^@AC^@ into three equal parts i.e ^@AC = 3AP = 3PQ = 3QC^@.
Answer:
- Through ^@ A ^@, let us draw ^@XAY || BC^@.
Now, ^@XY || MP || NQ^@ are cut by the transversal ^@AB^@ at ^@A, M,^@ and ^@ N ^@ respectively such that ^@AM = MN^@.
Also, ^@XY || MP || NQ^@ are cut by the transversal ^@AC^@ at ^@A, P, ^@ and ^@ Q ^@ respectively. @^ \begin{aligned} \therefore \space\space AP = PQ &&\ldots\text{(i)} &&[\text{ By intercept theorem }] \end{aligned} @^ - Again, ^@MP || NQ || BC^@ are cut by the transversal ^@AB^@ at ^@M, N,^@ and ^@ B^@ respectively such that ^@MN = NB^@.
Also, ^@MP || NQ || BC^@ are cut by the transversal ^@AC^@ at ^@P, Q,^@ and ^@ C^@ respectively. @^ \begin{aligned} \therefore \space\space PQ = QC &&\ldots\text{(ii)} &&[\text{ By intercept theorem }] \end{aligned} @^ Thus, from ^@eq \space (\text{i})^@ and ^@eq \space (\text{ii})^@, we get ^@AP = PQ = QC.^@
Therefore, ^@ P ^@ and ^@ Q ^@ divide ^@AC^@ into three equal parts.
Hence, ^@AC = 3AP = 3PQ = 3QC^@.
