Let ABC be the given triangle. Bisect ∠ABC and ∠BCA
by BD and CD (I.9), meeting at D. Drop perpendiculars DE,
DF, DG from D to AB, BC, CA (I.12). In the pairs of
triangles formed at D, the two angles and a common side give
congruence (I.26), whence DE=DF=DG. The circle with centre D
and radius DE touches each side (since the perpendiculars at the
feet make the sides tangents by III.16) and is inscribed in
△ABC.