In fig., l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. Another tangent at C intersects the line l at D and m at E. Prove that ∠DOE = 90°

Join OC
In triangle ODA and triangle ODC
A=OC (Radii of the same circle )
AD=DC (Length of tangent drawn from an external point to a circle are equal)
DO=OD (common side)
△DOA≅△ODC
∴∠DOA=∠COD
△DOA≅△ODC
∴∠DOA=∠COD
Similarly
△OEB≅△OEC
∴∠EOB=∠COE
AOB is a diameter of the circle.
Hence, it is a straight line.
∴∠DOA+∠COD+∠COE+∠EOB=180
⇒2∠COD+2∠COE=180
⇒∠COD+∠COE=90
⇒∠DOE=90°
Hence proved.