Question

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

Answer 1

Answer:

$\red{Given:-}$

$OA \times OB =OC \times OD$

$\red{To \:prove:-}$

$\angle A = \angle C\:\: and\:\:\angle B=\angle D$

$\red{Proof:-}$

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

$so,\angle A= \angle C \:and \: \: \angle D = \angle B$

$\orange{Hence\:\: Proved}$