# Existence of a triangle with sides partitioned in particular ratios by inscribed circle

Is there a triangle with sides that are partitioned into line segments of ratios $$3:2$$, $$3:5$$, and $$10:9$$ by the points of tangency of its inscribed circle?

By Ceva's Theorem, if a triangle has sides of lengths 20, 16, and 19, the cevian between the vertex and the point that partitions the side of length 20 into line segments of lengths 12 and 8, the cevian between the vertex and the point that partitions the side of length 16 into line segments of lengths 6 and 10, and the cevian between the vertex and the point that partitions the side of length 19 into line segments of lengths 10 and 9 coincide. Is this point the center of the inscribed circle?