62. Nice and easy problem with a right triangle.

by Virgil Nicula, Jul 14, 2010, 5:45 PM

Quote:
Let $ABC$ be a $A$-right triangle. Denote the midpoint $M$ of $AC$ and $D\in BC$ for which $AD\perp BC$ . Consider $P\in BC$ so that $MP\perp MB$ and $N\in AC$ so that $NP\perp NB$ , i.e. the quadrilateral $BNMC$ is inscribed in circle with diameter $[BP]$ . Prove that $R\in AD\cap BN$ is the midpoint of $[AD]$ .
Proof. Observe that $\widehat{ABN}\equiv\widehat {MNP}\equiv\widehat{MBP}$ . Therefore $\widehat{ABN}\equiv\widehat{MBC}$ , i.e. $[BN$ is $B$-symmedian in $\triangle ABC$ . Since $BAD\sim BCA$ obtain that $[BN$ is $B$-median in $\triangle BAD$ .
This post has been edited 7 times. Last edited by Virgil Nicula, Nov 23, 2015, 4:11 PM

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