Răspuns: a) AB=AC⇒∡ABC=∡ACB=(180°-∡BAC)÷2=75°
b) D-sim.(B,AC)⇒BD⊥AC si BM≡MD
∡AMB=∡AMD=90°
AM-lat.com.
de aici⇒ΔAMB≡AMD⇒AB=AD
∡BAM=∡DAM=30°
∡BAD=∡BAM+∡DAM=30°+30°=60°(1)
AB≡AD(2)
din (1) si (2)⇒ΔABD-echilateral
c)ΔBMC≡ΔDMC⇒BC≡CD
⇒(prin cazul L.L.L)ΔABC≡ΔADC
