Fie \( ABC \) un triunghi unghi ascutitunghic si fie \( D \) un punct in interiorul triunghiului astfel incat \( m(\angle ADB)-m(\angle ACB)=90^o \) si \( AC \cdot BD=AD \cdot BC \).
a) Sa se calculeze suma masurilor unghiurilor \( \angle DAC \) si \( \angle DBC \).
b) Sa se calculeze \( \frac{AB \cdot CD}{AC \cdot BD} \)
ONM problema 4
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Re: ONM problema 4
Frumoasa problema ! Mi-am permis sa extind neesential enuntul problemei propuse.
\( m(\angle BDE)=\phi \) , \( \frac {EB}{DB}=2\cdot \sin\frac {\phi}{2} \) si \( \triangle ACB\sim\triangle ADE \) deoarece \( \angle ACB\equiv\angle ADE \) si \( \frac {CA}{CB}=\frac {DA}{DB} \) . Rezulta
\( \angle BAC\equiv\angle EAD \) , \( \angle BAE\equiv\angle CAD \) si \( \frac {AB}{AE}=\frac {AC}{AD} \). Asadar, \( \triangle BAE\sim\triangle CAD \) .
In concluzie, \( \frac {AB}{AC}=\frac {EB}{DC} \) si \( \frac {AB}{AC}\cdot\frac{DC}{DB}=\frac {EB}{DC}\cdot\frac {DC}{DB}=\frac {EB}{DB}=2\cdot \sin\frac {\phi}{2} \) , adica \( \frac {AB}{AC}\cdot\frac{DC}{DB}=2\cdot \sin\frac {\phi}{2} \) .
Dem. Consideram punctul \( E \) interior unghiului \( \angle ADB \) pentru care \( m(\angle ADE)=C \) si \( DE=DB \) . Se observa caFie \( \triangle ABC \) ascutitunghic si un punct \( D \) interior acestuia ca \( \frac {DA}{DB}=\frac {CA}{CB} \)
si \( m\left(\angle ADB\right)=C+\phi\ <\ 180^{\circ} \) . Sa se arate ca \( \frac {AB}{AC}\cdot\frac{DC}{DB}=2\cdot \sin\frac {\phi}{2} \) .
\( m(\angle BDE)=\phi \) , \( \frac {EB}{DB}=2\cdot \sin\frac {\phi}{2} \) si \( \triangle ACB\sim\triangle ADE \) deoarece \( \angle ACB\equiv\angle ADE \) si \( \frac {CA}{CB}=\frac {DA}{DB} \) . Rezulta
\( \angle BAC\equiv\angle EAD \) , \( \angle BAE\equiv\angle CAD \) si \( \frac {AB}{AE}=\frac {AC}{AD} \). Asadar, \( \triangle BAE\sim\triangle CAD \) .
In concluzie, \( \frac {AB}{AC}=\frac {EB}{DC} \) si \( \frac {AB}{AC}\cdot\frac{DC}{DB}=\frac {EB}{DC}\cdot\frac {DC}{DB}=\frac {EB}{DB}=2\cdot \sin\frac {\phi}{2} \) , adica \( \frac {AB}{AC}\cdot\frac{DC}{DB}=2\cdot \sin\frac {\phi}{2} \) .