Given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there cannot be constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.
在同一底边、同一侧，不能另取一点，使从两端点引出的两线分别等于已有两线。
在底 AB 同侧假设存在两个不同的顶点 C 与 D，使得 AC=AD 与 BC=BD（反证后矛盾）。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
For, if possible, given two straight lines AC, CB constructed on the straight line AB and meeting at the point C, let two other straight lines AD, DB be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA is equal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it; and let CD be joined. Then, since AC is equal to AD, the angle ACD is also equal to the angle ADC; [I. 5] therefore the angle ADC is greater than the angle DCB; therefore the angle CDB is much greater than the angle DCB.
假设同底 AB、同侧有两个不同点 C、D，且 AC=AD、BC=BD。
Again, since CB is equal to DB, the angle CDB is also equal to the angle DCB.
连接 CD。由等腰三角形性质，围绕 CD 的相关底角会成对相等。
But it was also proved much greater than it: which is impossible.
这些角又互为包含关系，会推出较小角等于较大角。
Therefore etc.
矛盾来自假设有第二个点 D；所以这样的第二组线段不存在。
不看完整证明，说明本命题中为什么不能只凭图形直观看出结论。请至少提到一个前提和一个要证关系：在同一底边、同一侧，不能另取一点，使从两端点引出的两线分别等于已有两线。
看一个提示
- 先写“已知什么”，再写“要证什么”。
- 把你用到的定义、公设或前文命题编号写出来。