Asemanare intr-un pentagon.
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Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
Asemanare intr-un pentagon.
Se considera un pentagon convex \( ABCDE \) . Notam \( I\ \in\ BD\ \cap\ CE \) si circumcercurile \( \omega_1 \) , \( \omega_2 \) pentru \( \triangle ABC \) , \( \triangle ADE \) respectiv. Sa se arate ca \( I\in \omega_1\ \cap\ \omega_2\ \Longleftrightarrow\ \triangle ABC\ \sim\ \triangle ADE \) .
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Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
\( \triangle ABC\sim\triangle ADE\Longleftrightarrow \angle EAD=\angle BAC \) si \( \angle ADE=\angle BAC \)
Notand J intersectia celor doua cercuri deducem ca \( \angle EDA=\angle AJE \) si \( \angle ABC+\angle AJC=180^{\circ} \) de unde E, J si C sunt coliniare.
Deasemenea \( \angle DJE=\angle DAE=\angle BAC=\angle BJC \) deci si D,J si B sunt coliniare. Asadar I=J.
Reciproc , analog.
Notand J intersectia celor doua cercuri deducem ca \( \angle EDA=\angle AJE \) si \( \angle ABC+\angle AJC=180^{\circ} \) de unde E, J si C sunt coliniare.
Deasemenea \( \angle DJE=\angle DAE=\angle BAC=\angle BJC \) deci si D,J si B sunt coliniare. Asadar I=J.
Reciproc , analog.