clasele VI-VII, Lema 20-30

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Lema: Fie $\triangle ABC$ un triunghi cu $\angle B=30^{\circ}$ si $\angle C=20^{\circ}$. Fie un punct $D\in(BC)$. Aratati ca $BD=AC\Leftrightarrow DA=DC$.

solutia lemei o gasiti aici - http://forum.gil.ro/viewtopic.php?f=20&t=154

Aplicatii:
1. Fie un triunghi $\triangle ABC$ cu $m(\angle B)=100^{\circ}$. Fie $D\in(AB)$ astfel incat $AC=BD$. Calculati $\angle ACD$.

2. Fie un triunghi $\triangle BCD$ si $A$ un punct in interior astfel incat $\angle CBA=\angle ABD=20^{\circ}$, $\angle BCA=30^{\circ}$, $\angle DCA=10^{\circ}$. Calculati $\angle BDA$ si demonstrati ca $AB=BD$.

3. Fie $\triangle ABC$ un triunghi si $D$ un punct in interior astfel incat $\angle ACD=20^{\circ}$, $\angle DCB=40^{\circ}$, $\angle CBD=70^{\circ}$, $\angle DBA=10^{\circ}$. Demonstra\5i ca $AD\perp BC$.

4. In triunghiul $\triangle ABC$ se considera dreapta $BD$ cu $D\in(AC)$ astfel incat $AD=BC$. Daca $\angle A=30^{\circ}$ si $\angle C=40^{\circ}$, calculati $\angle DBC$.

5. Fie $\triangle ABC$ un triunghi isoscel cu $\angle B=\angle C=80^{\circ}$. $\{F\}=CF\cap AB$. $\angle ACF=30^{\circ}$. $\{E\}=AC\cap AB$. $m(\angle ABE)=20^{\circ}$. Aratati ca $\angle BEF=30^{\circ}$.