Question

use the diagram to prove the concurrency of angle bisectors theorem.
given: rays r, s, and t are bisectors of the angles of △abc, x is the intersection of rays r and s, ( overline{xd} perp overline{ac} ), ( overline{xe} perp overline{ab} ), and ( overline{xf} perp overline{bc} ).
prove: ray t contains point x, and ( xd = xe = xf )

statements

1. rays r, s, and t are bisectors of the angles of ( \triangle abc ), x is the intersection of rays r and s, ( overline{xd} perp overline{ac} ), ( overline{xe} perp overline{ab} ), and ( overline{xf} perp overline{bc} ).
2. ( xe = xd )
3. ( xe = xf )

reasons

1. given
2. ray r bisects ( angle bac ), so x is equidistant from the sides of ( angle bac ).
3. ray s bisects ( angle abc ), so x is

the sides of ( angle abc ).

Explanation:

For the third row under "Reasons" (corresponding to "Ray s bisects ∠ABC, so X is [blank] the sides of ∠ABC"), we use the Angle Bisector Theorem's property: a point on an angle bisector is equidistant from the angle's sides. So the missing word is "equidistant from" (matching the structure of the previous row where Ray r bisects ∠BAC and X is equidistant from its sides).

For the "Statements" column:

• Step 2: Since Ray r bisects ∠BAC, and XD ⊥ AC, XE ⊥ AB, by the Angle Bisector Theorem (a point on an angle bisector is equidistant from the sides of the angle), we conclude XE = XD.
• Step 3: Similarly, Ray s bisects ∠ABC, with XE ⊥ AB and XF ⊥ BC, so by the same theorem, XE = XF. Then, by the Transitive Property (if a = b and b = c, then a = c), since XE = XD and XE = XF, we get XD = XE = XF.

Answer:

Reasons (Third Row): equidistant from

Statements & Reasons Completion:

• Statement 2 (XE = XD) Reason: A point on an angle bisector is equidistant from the angle’s sides (Angle Bisector Theorem).
• Statement 3 (XE = XF) Reason: A point on an angle bisector is equidistant from the angle’s sides (Angle Bisector Theorem); then by Transitive Property, \( XD = XE = XF \).