Useful lemma
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Raiyan Jamil
- Posts:138
- Joined:Fri Mar 29, 2013 3:49 pm
In triangle $ABC$ the $incircle$ touches $BC$,$CA$ and $AB$ at $D$,$E$ and $F$ respectively. Let the mid-points of $BC$,$CA$ and $AB$ be $M$,$N$ and $P$ respectively. Prove that, $1)$ $AI$,$DE$,$MP$ and the perpendicular from $B$ to $AI$ are concurrent. $2)$ $AI$,$DF$,$MN$ and the perpendicular from $C$ to $AI$ are concurrent.
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Re: Useful lemma
The proofs of (1) and (2) are same.So,let's prove (1).
Let $AI \cap DE=Q,DE \cap AB=R$.$(B,A;F,R)=-1$.Now $\triangle AFQ \cong \triangle AEQ$ $\Rightarrow$
$\angle FQA=\angle AQE$.So,$\angle AQB=90^{\circ}$.Now,$\angle PQA=\angle PAQ=\angle QAC$
$\Rightarrow$ $PQ \parallel AC$ $\Rightarrow$ $PQ$ passes through $M$.So,$AI,DE,MP$ and the perpendicular from $B$ to $AI$ are concurrent.
(2) can be done similarly.
Let $AI \cap DE=Q,DE \cap AB=R$.$(B,A;F,R)=-1$.Now $\triangle AFQ \cong \triangle AEQ$ $\Rightarrow$
$\angle FQA=\angle AQE$.So,$\angle AQB=90^{\circ}$.Now,$\angle PQA=\angle PAQ=\angle QAC$
$\Rightarrow$ $PQ \parallel AC$ $\Rightarrow$ $PQ$ passes through $M$.So,$AI,DE,MP$ and the perpendicular from $B$ to $AI$ are concurrent.
(2) can be done similarly.
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