Question

In $\triangle PQR$, seg $PM$ is a median. Angle bisectors of $\angle PMQ$ and $\angle PMR$ intersect sides $PQ$ and $PR$ in points $X$ and $Y$ respectively. Prove that $XY \parallel QR$. 

Hint:

When medians and angle bisectors are involved in triangles, use the Angle Bisector Theorem and the Converse of the Basic Proportionality Theorem to prove parallel lines.

Answer 1

Step 1: In $\triangle PMQ$,
Ray $MX$ is the bisector of $\angle PMQ$.
By the Angle Bisector Theorem, \[ \frac{MP}{MQ} = \frac{PX}{XQ} \quad \text{.............. (I)} \] Step 2: In $\triangle PMR$,
Ray $MY$ is the bisector of $\angle PMR$.
By the Angle Bisector Theorem, \[ \frac{MP}{MR} = \frac{PY}{YR} \quad \text{.............. (II)} \] Step 3: Since $M$ is the midpoint of $QR$,
\[ MQ = MR \] Hence, \[ \frac{MP}{MQ} = \frac{MP}{MR} \quad \text{.............. (III)} \] Step 4: From (I), (II), and (III),
We get, \[ \frac{PX}{XQ} = \frac{PY}{YR} \] Step 5: By the Converse of the Basic Proportionality Theorem (Thales Theorem),
\[ XY \parallel QR \] Hence, it is proved that $XY \parallel QR$.