Question

use the diagram to prove the concurrency of angle bisectors theorem.
given: 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 $\overline{xf}\perp\overline{bc}$
prove: ray t contains point x, and $xd = xe = xf$

supply a reason for each step of the following proof.

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 of $\angle bac$
3. ray s bisects $\angle abc$, so x is the sides of $\angle abc$

(options: equidistant from, congruent to, between, congruent to)

Explanation:

1. For step 2: A point on an angle bisector is equidistant from the angle's sides. Since ray r bisects ∠BAC, X (on r) is equidistant from AB and AC, so XE = XD.
2. For step 3: A point on an angle bisector is equidistant from the angle's sides. Since ray s bisects ∠ABC, X (on s) is equidistant from AB and BC, so XE = XF.
3. By transitivity, XD = XE = XF. Since X is equidistant from AC and BC, it lies on the angle bisector of ∠ACB, which is ray t, so ray t contains X.

Answer:

Completed Proof Reasons:

1. Given
2. equidistant from
3. equidistant from

Full Logical Conclusion:

Ray \( t \) contains point \( X \), and \( XD = XE = XF \)