192. An easy and nice problem of Valentin Vornicu.

by Virgil Nicula, Dec 18, 2010, 8:16 AM

Proposed problem. Let $ABC$ be an $A$-isosceles triangle. Denote the midpoint $ M$ of the side $[BC]$ and the point $ O\in AM$ such that $ OB\perp AB$. Consider

an arbitrary point $Q\in BC$ different from $ B$ and $ C$ and the points $ E\in AB\ ,\ F\in AC$ such that $Q\in EF$ . Prove that $ OQ\perp EF\ \iff\ QE=QF$.

Proof 1. Denote the reflection $E'$ of the point $E$ w.r.t. the point $B$ , i.e. $BO$ is the bisector of the segment $[EE']$ . From the Menelaus' theorem applied to the transversal

$\overline {BQC}$ for $\triangle AEF$ results the relation $\frac{BE}{BA}\cdot \frac{CA}{CF}\cdot \frac{QF}{QE}=1$ , i.e. $\frac{BE'}{CF}=\frac{QE}{QF}$ . Thus, $QE=QF\Longleftrightarrow BE'=CF\Longleftrightarrow E'F\parallel BC\Longleftrightarrow$

$AO$ is the bisector of the segment $[E'F]\Longleftrightarrow$ the point $O$ is the circumcenter of the triangle $E'EF\Longleftrightarrow OQ\perp EF$ .

Proof 2. Supppose w.l.,o.g. $Q\in (MC)$ . Observe that $OB=OC$ and $OQ\perp EF\iff$ $BEQO$ and $CFOQ$ are cyclically $\iff$

$\widehat {OEF}\equiv$ $\widehat {OEQ}\equiv$ $\widehat {OBQ}\equiv$ $\widehat {OBC}\equiv$ $\widehat {OCB}\equiv$ $\widehat {OCQ}\equiv$ $\widehat {OFQ}\equiv$ $\widehat {OFE}\iff$ $\widehat {OEF}\equiv\widehat {OFE}$ $\iff$ $QE=QF$ .

Quote:
An easy extension. Let $w$ be the circumcircle of $\triangle ABC$ . Let $[AA']$ be the diameter of $w$ . Consider the points

$ D\in BC\ ,\ E\in CA\ ,\ F\in AB$ so that $D\in EF$ . Prove that $A'D\perp EF\ \iff\ \frac {A'E}{A'F} = \frac {A'C}{A'B}$ .

Proof. $==\blacktriangleright$ Suppose that $A'D\perp EF$ . Then $ A'B\perp AB$ $\implies$ $ BDA'F$ is cyclically $\implies$ $\widehat{EFA'}\equiv\widehat{DFA'}\equiv\widehat{DBA'}\equiv\widehat{CBA'}\equiv$

$\widehat{CAA'}\equiv\widehat{EAA'}$ $\implies$ $(1)\ \boxed{\widehat{CBA'}\equiv\widehat{EFA'}}\equiv\widehat{EAA'}$ $\implies$ $AEA'F$ is cyclically $\implies$ $\widehat{BCA'}\equiv\widehat{BAA'}\equiv$ $\widehat{FAA'}\equiv\widehat{FEA'}$ $\implies$

$(2)\ \boxed{\widehat{BCA'}\equiv\widehat{FEA'}}$ . From the relations $(1)$ and $(2)$ obtain that $\triangle EA'F\sim\triangle CA'B\ \implies\ \frac{A'F}{A'E}=\frac{A'B}{A'C}$ .

$\blacktriangleleft ==$ Suppose that $ \frac{A'F}{A'E}=\frac{A'B}{A'C}$ . Then $\frac{A'F}{A'B}=\frac{A'E}{A'C}$ and $ \angle A'CE=\angle A'BF=\frac{\pi}{2}$ $\implies$ $ \angle A'FA=\angle A'EC$ $\implies$ $ AEA'F$ is

cyclically. Therefore $ \angle A'BC=\angle EAA'=\angle EFA'$ $\implies$ $ BDA'F$ is cyclically $\implies$ $ \angle FDA'=\frac{\pi}{2}$ $\implies$ $A'D\perp EF$ .

See and here CLVI - message of My (math)blog.

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This post has been edited 26 times. Last edited by Virgil Nicula, Nov 22, 2015, 5:34 PM

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