Date Example 6: In fig. 6.31. OA. 0D = 0C.0D show that LA=LC. LB = Le

1.	Date Example 6: In fig. 6.31. OA. 0D = 0C.0D show that LA=LC. LB = Le and LD
Answer» Answer: $\red{Given:-}$ $<klux>OA</klux> \times OB =<klux>OC</klux> \times OD$ $\red{To \:prove:-}$ $\angle A = \angle C\:\: and\:\:\angle B=\angle D$ $\red{Proof:-}$ OA . OB = OC . OD $\frac{OA}{OC} = \frac{OD}{OB} \:\:...(1)$ $In\: \triangle AOD \: and\: \triangle COB$ $\frac{OA}{OC} = \frac{OD}{OB} \: \: \: \: \: \:(From(1))$ $\angle AOD=\angle COB \: \: \: \: \: \: \: \: \: \: \: \:(vertically\;opposite \:angle)$ Using SAS SIMILARITY criterion $So,\triangle AOD \approx \triangle COB$ We know that if two triangles are similar so,their corresponding angles are equal $so,\angle A= \angle C \:and \: \: \angle D = \angle B$ $\orange{Hence\:\: Proved}$