若△ABC∽△A′B′C′,且字母绕向A→B→C与字母绕向A′→B′→C′相同(即同为顺时针方向或同为逆时针方向),我们称△ABC与△A′B′C′为同向相似.一、命题及其证明同向相似的两个三角形的对应边所在直线的夹角相等.已知:如图1,△ABC与△A′B′C′为同向相似.求证:AB与A′B′的夹角=BC If △ ABC ∽ △ A’B’C ’, and the letters around A → B → C are the same as letters around A’ → B ’→ C’ (ie the same clockwise or counterclockwise), we △ ABC and △ A’B’C ’for the same direction. First, the proposition and its proof with the same direction of the two triangles corresponding to the same angle of the same straight line is known: As shown in Figure 1, △ ABC and △ A’B’C ’for the same direction. Proof: AB and A’B’ angle = BC