L and M are two parallel tangents to a circle with center 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 angle DOE is 90 degrees?

Answer 1

Extend DO to meet M at F, then:

∠ADO = ∠OFB (alternate angles as DOF is a straight line across the two parallel lines)

Using the straight line FOD:

∠DOE = ∠BEO + ∠OFB (exterior angle of a triangle = sum of other two angles)

= ∠BEO + ∠ADO (substituting for equivalent angle found above)

Consider triangles OBE and OCE.

They are both right angle triangles as BE and CE are tangent to the circle centre O, with ∠OBE = ∠OEC = 90°.

They have the same hypotenuse side OE.

They have equivalent sides OB and OC equal in length (both are a radius of the circle).

Thus triangles OBE and OCE are congruent by Right Angle-Hypotenuse-Side (RHS).

Making ∠CEO = ∠BEO

Similarly considering the triangles OAD and OCD:

Both right angled triangles with ∠OAD = ∠OCD = 90°

Both have same hypotenuse OD

Both have equivalent side OA and OC of equal length (both are a radius of the circle).

Thus triangles OAD and OCD are congruent by RHS

Making ∠CDO = ∠ADO

Summing the angles in triangle ODE and substituting equivalent angles (∠CDO = ∠ADO, ∠CEO = ∠BEO, ∠DOE = ∠BEO + ∠ADO) gives:

∠CEO + ∠CDO + ∠DOE = 180°

→∠BEO + ∠ADO + ∠DOE = 180°

→ ∠DOE + ∠DOE = 180°

→ 2∠DOE = 180°

→ ∠DOE = 90°

QED.