Question

(02.03 MC) triangle ADB, point C lies on segment AB and forms segment CD, angle ACD measures 90 degrees. Point A is labeled jungle gym and point B is labeled monkey bars. Beth is planning a playground and has decided to place the swings in such a way that they are the same distance from the jungle gym and the monkey bars. If Beth places the swings at point D, how could she prove that point D is equidistant from the jungle gym and monkey bars? If segment DC bisects segment AB, then point D is equidistant from points A and B because a point on a perpendicular bisector is equidistant from the endpoints of the segment it intersects. If segment DC bisects segment AB, then point D is equidistant from points A and B because congruent parts of congruent triangles are congruent. If segment AD bisects segment AB, then point D is equidistant from points A and B because a point on a perpendicular bisector is equidistant from the endpoints of the segment it intersects. If segment AD bisects segment AB, then point D is equidistant from points A and B because congruent parts of congruent triangles are congruent.

Answer 1

Answer:

If segment DC bisects segment AB, then point D is equidistant from points A and B because congruent parts of congruent triangles are congruent.

Step-by-step explanation:

To prove that AD (distance from jungle gym to D) = BD (distance from monkey bars to D)

Given: <ACD = $90^{0}$

               CD, common side of ΔACD and ΔBCD

Then;

AC = BC (midpoint of /AB/)

<ACD = <BCD (right angle property)

ΔACD ≅ ΔBCD (congruent property of similar triangles)

Therefore by congruent property of Side-Angle-Side (SAS), /AD/ = /BD/. Because congruent parts of congruent triangles are equal.

Answer 2

Answer:

The answer above is wrong on FLVS

Step-by-step explanation: