Prove that $PQ \perp EF$.

$$I$$ is the incenter of $$\triangle ABC$$. The incircle of $$\triangle ABC$$ is tangent to $$BC$$, $$CA$$ and $$AB$$ respectively at points $$D$$, $$E$$ and $$F$$. Given that $$BM \perp DE$$ and $$CN \perp DF$$, $$(M \in DE$$ and $$N \in DF)$$. Construct points $$P$$ and $$Q$$ such that $$P$$ and $$Q$$ are respectively the midpoints of $$ID$$ and $$MN$$. Prove that $$PQ \perp EF$$.

• You should write clearly, It was come from HSGS Class 9 Contests! – Tran Quang Hung Jun 22 '19 at 11:23

Let $$BI \cap DF = K$$ and $$CI \cap DE = L$$.

$$\implies K$$ and $$L$$ are respectively the midpoints of $$DE$$ and $$DF$$, $$BK \perp DF$$ and $$CL \perp DE$$.

\implies \left\{ \begin{align} KL &\parallel FE\\ BK &\parallel CN\\ CL &\parallel BM \end{align} \right.\implies \left\{ \begin{align} \widehat{DKL} &= \widehat{DFE}\\ \widehat{DLK} &= \widehat{DEF}\\ \widehat{KBD} &= \widehat{NCD}\\ \widehat{LCD} &= \widehat{MBD}\\ \end{align} \right.

Furthermore, we have that $$\widehat{BKD} = \widehat{CLD} = \widehat{BMD} = \widehat{CND} \ (= 90^\circ)$$

$$\implies BKMD$$ and $$CLNC$$ are cyclic quadrilaterals.

\implies \left\{ \begin{align} \widehat{MBD} &= \widehat{MKD}\\ \widehat{NCD} &= \widehat{NLD} \end{align} \right.\implies \left\{ \begin{align} \widehat{KBD} &= \widehat{NLD}\\ \widehat{LCD} &= \widehat{MKD} \end{align} \right. (1)

In addition, $$BC$$ is a tangent of (I) \implies \left\{ \begin{align} \widehat{BDK} = \widehat{DEF}\\ \widehat{CDL} = \widehat{DFE} \end{align} \right.\implies \left\{ \begin{align} \widehat{DLK} = \widehat{BDK}\\ \widehat{DKL} = \widehat{CDL} \end{align} \right. (2)

From $$(1)$$ and $$(2)$$, we have that \left\{ \begin{align} \widehat{KBD} + \widehat{BDK} = \widehat{NLD} + \widehat{DLK} = \widehat{NLK}\\ \widehat{LCD} + \widehat{CDL} = \widehat{MKD} + \widehat{DKL} = \widehat{MKL}\end{align} \right.

Moreover, $$\widehat{KBD} + \widehat{BDK} = \widehat{LCD} + \widehat{CDL} = 90^\circ \implies \widehat{NLK} = \widehat{MKL} = 90^\circ$$

$$\implies MKLN$$ is a right-angled parallelogram at $$K$$ and $$L$$, it is also known that $$Q$$ is the midpoint of $$MN$$.

$$\implies QK = QL$$.

By the same token, looking at right-angled triangles $$IKD$$ and $$ILD$$ respectively at $$K$$ and $$L$$ where $$P$$ is the midpoint of $$ID$$, it can be seen that $$KP = LP \ \left(= \dfrac{ID}{2}\right)$$.

$$\implies PQ$$ is the perpendicular bisector of $$KL \implies PQ \implies KL \perp EF$$.