Math Question

draws \( \triangle K K L \), where \( \overline{J K} \perp \overline{K L} \) and \( \overline{K M} \perp \overline{J L} \).
Complete the proof that \( J K^{2}+K L^{2}=J L^{2} \).
You know that \( \overline{J K} \perp \overline{K L} \) and \( \overline{K M} \perp \overline{J L} \). Since perpendicular lines form right angles and all right angles are congruent, \( \angle J M K \cong \angle L M K- \). Also, by the Reflexive Property of Congruence, \( \angle J \cong \angle J \). So, \( \triangle J K L \sim \triangle M M K \) by the Angle-Angle \( \quad \) Similarity Theorem. Therefore, \( \frac{J K}{J L}=\frac{J M}{J K} \) because corresponding sides in similar figures are proportional. You can rewrite this equation as \( J K^{2}= \)
Similarly, since \( \overline{J K} \perp \overline{K L} \), and \( \overline{K M} \perp \overline{J L}, \angle J K L \cong \angle K M L \). Also, \( \angle L \cong \angle L \quad \) by the Reflexive Property of Congruence, and so, by the Angle-Angle Similarity Theorem, \( \triangle J K L \sim \)
Therefore, \( \frac{J L}{K L}= \)



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