Question

Any point X inside triangle DEF is joined to its vertices. From a point P on DX, PQ is drawn parallel to DE, meeting XE at Q and QR is drawn parallel to EF, meeting XF in R. Prove that PR is parallel to DF.

Answer 1

Answer on Question #43538 – Math - Analytic Geometry

Any point $X$ inside triangle DEF is joined to its vertices. From a point $P$ on DX, PQ is drawn parallel to DE, meeting XE at Q and QR is drawn parallel to EF, meeting XF in R. Prove that PR is parallel to DF.

Solution:

Let $\overline{XD} = \vec{a}$, $\overline{XE} = \vec{b}$, $\overline{XF} = \vec{c}$. Then $\overline{DE} = \vec{b} - \vec{a}$, $\overline{EF} = \vec{c} - \vec{b}$, $\overline{FD} = \vec{a} - \vec{c}$. Suppose that $\overline{XP} = \lambda \overline{XD} = \lambda \vec{a}$. The vector $\overline{PQ}$ is colliner to $\overline{DE}$, so $\overline{PQ} = \mu_1 \overline{DE} = \mu_1 (\vec{b} - \vec{a})$. Then $\overline{XQ} = \overline{XP} + \overline{PQ} = (\lambda - \mu_1) \vec{a} + \mu_1 \vec{b}$. On the other hand $\overline{XQ} = \mu_2 \overline{XE} = \mu_2 \vec{b}$, so $(\mu_1 - \lambda) \vec{a} + \mu_1 \vec{b} = \mu_2 \vec{b}$. The vectors $\vec{a}$ and $\vec{b}$ are linear independent, so $\mu_1 = \lambda$, $\mu_2 = \mu_1 = \lambda$ and $\overline{XQ} = \lambda \vec{b}$. The vector $\overline{QR}$ is colliner to $\overline{EF}$, so $\overline{QR} = \mu_3 \overline{EF} = \mu_3 (\vec{c} - \vec{b})$. Then $\overline{XR} = \overline{XQ} + \overline{QR} = (\lambda - \mu_3) \vec{b} + \mu_3 \vec{c}$. On the other hand $\overline{XR} = \mu_4 \overline{XF} = \mu_4 \vec{c}$, so $(\lambda - \mu_3) \vec{b} + \mu_3 \vec{c} = \mu_4 \vec{c}$. The vectors $\vec{b}$ and $\vec{c}$ are linear independent, so $\mu_3 = \lambda$, $\mu_4 = \mu_3 = \lambda$ and $\overline{XR} = \lambda \vec{c}$. Hence $\overline{RP} = \overline{XP} - \overline{XR} = \lambda (\vec{a} - \vec{c}) = \lambda \overline{FD}$, so vector $\overline{RP}$ is parallel to $\overline{FD}$ or $RP||FD$.

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