1. **Problem Statement:** We are given a right triangle $\triangle ABC$ with a right angle at $A$. Points $K$ and $Q$ lie on segments $AB$ and $AC$ respectively. Segment $BK$ is given as 23 units, and $BK$ is congruent to $QP$. We need to find the length of $KQ$. 2. **Understanding the Problem:** Since $BK$ and $QP$ are congruent, $BK = QP = 23$. The problem asks for $KQ$, which is the segment connecting points $K$ and $Q$ on sides $AB$ and $AC$. 3. **Key Geometry Concepts:** In a right triangle, the median to the hypotenuse is half the length of the hypotenuse. Also, if $K$ and $Q$ are points on legs $AB$ and $AC$, and $BK = QP$, then $KQ$ can be found using the properties of medians or altitudes depending on the exact configuration. 4. **Assuming $K$ and $Q$ are midpoints:** Since $BK$ and $QP$ are congruent and $K$ and $Q$ lie on $AB$ and $AC$, it is likely that $K$ and $Q$ are midpoints of $AB$ and $AC$ respectively. 5. **Using the Midpoint Theorem:** The segment $KQ$ connecting midpoints of two sides of a triangle is parallel to the third side and half its length. 6. **Calculate $KQ$:** Since $K$ and $Q$ are midpoints, $KQ = \frac{1}{2} BC$. 7. **Find $BC$ using Pythagoras:** Since $\triangle ABC$ is right angled at $A$, $$ BC = \sqrt{AB^2 + AC^2} $$ 8. **Given $BK = 23$ and $K$ is midpoint of $AB$, then $AB = 2 \times BK = 46$** 9. **Similarly, if $Q$ is midpoint of $AC$, and $QP = BK = 23$, then $AC = 2 \times QP = 46$** 10. **Calculate $BC$:** $$ BC = \sqrt{46^2 + 46^2} = \sqrt{2 \times 46^2} = 46 \sqrt{2} $$ 11. **Calculate $KQ$:** $$ KQ = \frac{1}{2} BC = \frac{1}{2} \times 46 \sqrt{2} = 23 \sqrt{2} $$ **Final answer:** $$ KQ = 23 \sqrt{2} $$