1. **Problem 11:** In triangle ABC, given AB = AC and points X on AB and Y on AC such that AX = AY, prove that \(\triangle ABY \cong \triangle ACX\). 2. **Formula and rule:** To prove two triangles congruent, we can use the SAS (Side-Angle-Side) criterion: if two sides and the included angle of one triangle are equal to the corresponding parts of another triangle, the triangles are congruent. 3. **Step-by-step proof:** - Given: \(AB = AC\), \(AX = AY\). - Since X lies on AB and Y lies on AC, segments BX and CY are parts of AB and AC respectively. - Angle \(\angle BAX = \angle CAY\) because they are the same angle \(\angle BAC\). - In \(\triangle ABY\) and \(\triangle ACX\), we have: - Side \(AB = AC\) (given) - Side \(AY = AX\) (given) - Angle \(\angle BAX = \angle CAY\) (common angle) - By SAS criterion, \(\triangle ABY \cong \triangle ACX\). --- 4. **Problem 12:** In isosceles triangle LMN with \(LM = LN\), LO bisects \(\angle MLN\). Prove O is midpoint of MN. 5. **Formula and rule:** In an isosceles triangle, the angle bisector of the vertex angle also bisects the base. 6. **Step-by-step proof:** - Given \(LM = LN\) and LO bisects \(\angle MLN\). - Triangles \(\triangle LMO\) and \(\triangle LNO\) share side LO. - Angles \(\angle MLO = \angle NLO\) (LO bisects \(\angle MLN\)). - Sides \(LM = LN\) (given). - By SAS, \(\triangle LMO \cong \triangle LNO\). - Corresponding parts of congruent triangles are equal, so \(MO = ON\). - Hence, O is midpoint of MN. --- 7. **Problem 13:** In triangle with \(LM = MN\), \(QM = MR\), \(ML \perp PQ\), and \(MN \perp PR\), prove \(PQ = PR\). 8. **Formula and rule:** Use congruency criteria and properties of perpendicular lines. 9. **Step-by-step proof:** - Given \(LM = MN\), \(QM = MR\), \(ML \perp PQ\), \(MN \perp PR\). - Consider triangles \(\triangle QML\) and \(\triangle RMP\). - \(LM = MN\) (given). - \(QM = MR\) (given). - Angles \(\angle QML = \angle RMP = 90^\circ\) (given perpendiculars). - By RHS (Right angle-Hypotenuse-Side) congruency, \(\triangle QML \cong \triangle RMP\). - Corresponding parts are equal, so \(PQ = PR\). --- **Final answers:** - \(\triangle ABY \cong \triangle ACX\) - O is midpoint of MN - \(PQ = PR\)