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In triunghiul \( ABC \), \( m(\angle A)=90^\circ \), \( [AD] \) este inaltime si \( M \in (AD) \). Sa se arate ca daca \( \angle MBA \equiv \angle MCA \) atunci \( AB=AC \).
Manuela Prajea, R.M.T. 4/2008

\( \frac{A[ABM]}{A[ACM]}=\frac{AB\cdot BM\cdot \sin{\angle{MBA}}}{AC\cdot CM\cdot \sin{\angle{MCA}}}=\frac{BD\cdot AM}{CD\cdot AM}=\frac{AB^2}{AC^2} \) si de aici

\( \frac{BD}{CD}=\frac{AB}{AC} \)

Apoi cu putine calcule rezulta ca BD=CD deci triunghiul ABC este isoscel.

Solutia mea.
Fie \( { N} =BM \cap AC, {P}= MC \cap AB \).
Deoarece \( \angle MBA \equiv \angle MCA \Longrightarrow \angle PBN \equiv \angle PCN \), deci patrulaterul \( PNCB \) este inscriptibil, sau \( PN \) este antiparalela la \( BC \), deci \( \Delta ANP \sim \Delta ABC\Longrightarrow \frac{AN}{AB}=\frac{AP}{AC}.(1) \), deci \( AN=\frac{AP\cdot AB}{AC} \) si \( AP=\frac{AN \cdot AC}{AB} \)

Dar cu teorema lui Ceva, avem: \( \frac{AP}{PB}\cdot \frac{BD}{DC} \cdot \frac{CN}{NA}=1 \Longleftrightarrow \frac{AP}{PB} \cdot \frac{CN}{NA} \cdot \frac{AB^2}{AC^2}=1\Longleftrightarrow \frac{AN \cdot AC}{AB \cdot PB} \cdot \frac{CN \cdot AC}{AP \cdot AB}\cdot \frac{AB^2}{AC^2}=1\Longleftrightarrow \frac{AN}{PB}=\frac{AP}{CN}\Longleftrightarrow \frac{AN}{AP}=\frac{PB}{CN}.(*) \).

Apoi, \( \Delta ABN \sim \Delta ACP \Longleftrightarrow \frac{AN}{AP}=\frac{AB}{AC}(**) \).
Din \( (*) \) si \( (**) \) rezulta ca \( \frac{PB}{CN}=\frac{AB}{AC} \Longleftrightarrow \frac{PB}{AB}=\frac{CN}{AC} \), ceea ce in conformitate cu reciproca teoremei lui Thales inseamna \( PN \parallel BC \) si deci \( \angle ANP =\angle ABC \) si \( \angle ANP=\angle ACB \), deci \( \angle ABC=\angle ACB \Longleftrightarrow AB=AC \qed \)