The incircle of △A0B0C0, meets legs B0C0, C0A0, A0B0, respectively on points A, B, C, and the incircle of △ABC, with center I, meets legs BC, CA, AB, on points A1, B1, C1, respectively. We write with σ(ABC), and σ(A1B1C1) the areas of △ABC, and △A1B1C1 respectively. Prove that if σ(ABC)=2σ(A1B1C1), then lines AA0, BB0, CC0 are concurrent.