Răspuns :
ΔABC dr.
m(∡B)=90°
BD=mediana⇔D mijl. (AC)⇒BD=AC/2
AC=AD+DC
=2DC ⇒BD=DC⇒ΔBDC isoscel
m(∡C)=60° ⇒
ΔBDC echilateral
m(∡B)=90°
BD=mediana⇔D mijl. (AC)⇒BD=AC/2
AC=AD+DC
=2DC ⇒BD=DC⇒ΔBDC isoscel
m(∡C)=60° ⇒
ΔBDC echilateral
Teorema medianei ⇒BD≡DC⇒ ΔBDC-ISOSCEL ∡C=60,ΔBDC ISOSCEL
⇒ΔBDC -echilateral
⇒ΔBDC -echilateral