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$I$ 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 $PQ \perp EF$.

It's sad that my solution doesn't follow the shortcut. (There's multiple.)

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

enter image description here 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 $KL \implies PQ \implies KL \perp EF$.


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