Question 42

1 2 + = . OA OB OC Solution: In ∆ AOF and ∆ BOD. ∠O = ∠O (Same angle) and ∠A = ∠B (each 90°) Therefore, ∆ AOF ~ ∆ BOD (AA similarity) OA FA So, = (1) OB DB Also, in ∆ FAC and ∆ EBC, ∠A = ∠B (Each 90°) and ∠FCA = ∠ECB (Vertically opposite angles). Therefore, ∆ FAC ~ ∆ EBC (AA similarity). FA AC So, = EB BC But EB = DB (B is mid-point of DE) FA AC So, = (2) DB BC Therefore, from (1) and (2), we have: AC OA = BC OB OC–OA OA i.e., = OB–OC OB or OB . OC – OA . OB = OA . OB – OA . OC or OB . OC + OA . OC = 2 OA . OB or (OB + OA). OC = 2 OA . OB