Răspuns: a)AP≡AB
∡ABC=∡APD=90°(PD⊥AC)
∡BAC≡∡DAP
de aici ⇒(prin cazul L.U.L)ΔABC≡ΔAPD
b) ΔABC≡ΔAPD⇒AC≡AD
∡BAC=60°
de aici ⇒ΔACD-echilateral
c)CB⊥AD⇒CB-inaltime in ΔACD
PD⊥AC⇒PD-inaltime in ΔACD
BC∩PD={H}
de aici⇒H-ortocentrul ΔACD⇒AH⊥CD
sper ca am ajuat!
