Question

Let M be the midpoint of side \overline{AB} of \triangle ABC. Angle bisector \overline{AD} of \angle CAB and the perpendicular bisector of side \overline{AB} meet at X. If AB = 40 and MX = 9, then how far is X from line {AC}?

Answer 1

To find the distance from X to line AC, we can use the angle bisector theorem and the fact that M is the midpoint of AB. By substituting the known values, we can solve for the distance d and find that X is 12 units away from line AC.

Step-by-step explanation:

We are given that M is the midpoint of side AB of triangle ABC. Angle bisector AD of angle CAB and the perpendicular bisector of side AB meet at X. We are also given that AB = 40 and MX = 9.

To find the distance from X to line AC, we can use similar triangles. Let's denote the distance from X to line AC as d. According to the angle bisector theorem, we have:

AD/CD = AB/CB

Since M is the midpoint of AB, we have:

MD = MB = AB/2 = 40/2 = 20

Therefore, we can rewrite the angle bisector theorem as:

AD/(AD + CD) = AB/CB

Substituting the known values, we get:

9/(9 + d) = 40/20

Cross multiplying, we have:

20 * 9 = 40 * (9 + d)

Simplifying, we find:

d = 12

Therefore, X is 12 units away from line AC.

Answer 2

Answer: 9

Step-by-step explanation:

According to the statement we have

AD is bisecting the angle <CAB

and Perpendicular bisector of overline AB i.e. CM is meeting at X.

Now, AB=40 and MX= 9.

We can already relate that the two triangles

AMX and AEX are congruent that means,

the ratio of sides are equal i.e. AM/EX= AM/MX

or EX= MX

Now, MX= 9 so EX=9 which is actually the distance of X from line AC