Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 5 - Relationships Within Triangles - 5-2 Perpendicular and Angle Bisectors - Practice and Problem-Solving Exercises - Page 296: 15


$m \angle KHL$ is $54^{\circ}$. $m \angle KHL$ is $54^{\circ}$.

Work Step by Step

The converse of the angle bisector theorem states that a point inside an angle that is equidistant from the two sides of the angle is located on the bisector of that angle. In this diagram, we see that the ray $HL$ is equidistant from the two sides of $\angle KHF$, so ray $HL$ is the angle bisector of $\angle KHF$. Therefore, $\angle KHL$ and $\angle FHL$ are congruent. So we can set the two angles equal to one another to solve for $y$: $6y = 4y + 18$ Subtract $4y$ from each side of the equation to isolate the variable on the left side of the equation: $2y = 18$ Divide each side of the equation by $2$ to solve for $y$: $y = 9$ Now that we have the value for $y$, we can substitute it into the expression for one of the angles because the angles are the same: $m \angle KHL = 6y$ Substitute $9$ for $y$: $m \angle KHL = 6(9)$ Solve by multiplying: $m \angle KHL = 54^{\circ}$ If $m \angle KHL$ is $54^{\circ}$, then $m \angle KHL$ is also $54^{\circ}$.
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