In ΔABC, B is a right angle, and BD is perpendicular to AC and DE is perpendicular to BC. Then AD.DC =
BD2
BE.BC
In ΔABD & ΔBDC∠ADB=∠BDC=90∘∠A=∠DBC ( Since in ΔBDC,∠DBC=90∘−∠C=∠A
In ΔABC∠A=90∘−∠C)∴,ΔADB∼ΔBDC by AA-similarity
ADDB=BDDC=ABBC⇒BD2=AD.DC
Similarly ΔBDC∼ΔBED∠DBC=∠DBE (Common angle)∠BDC=∠DEB = 90∘BDBE=DCED=BCBD⇒BD2=BE.BC⇒BD2=AD.DC=BE.BC