Math Help - Computing the distance

1. Computing the distance

Triangle ABC has side lengths AB = 231;BC = 160; and AC = 281: Point D is constructed on
the opposite side of line AC as point B such that AD = 178 and CD = 153: How to compute the distance
from B to the midpoint of segment AD

2. Re: Computing the distance

Hello, Mhmh96!

I assume you made a sketch . . .

Triangle ABC has side lengths AB = 231, BC = 160, and AC = 281.
Point D is constructed on the opposite side of line AC as point B such that AD = 178 and CD = 153.
Compute the distance from B to the midpoint of segment AD.

The diagram seems to be something like this.
Code:
                A
o     178
*  *   *      D
*     *       o
*        *      *
231 *           *     *
*          281 *    *
*                 *   * 153
*                    *  *
*                       * *
*                          **
B o  *  *  *  *  *  *  *  *  *  o C
160

We find that: . $160^2 + 231^2 \:=\:281^2$
. . Hence: . $\angle B \,=\,90^o.$

Now the diagram looks like this:
Code:
    A o
* **
* 1 * * 178
*     *  *
*       *   *
*         *    *   D
231 *           *     o
*         281 *    *
*               *   *
*                 * 2* 153
*                   * *
*                     **
B o  *  *  *  *  *  *  *  o C
160
Let $\angle 1 = \angle BAC,\;\angle2 = \angle ACD$

Law of Cosines:

$\cos(\angle 1) \:=\:\frac{231^2 + 281^2 - 160^2}{2(231)(281)} \:=\:\frac{106,\!722}{129,\!822} \:=\:\frac{231}{281}$

$\cos(\angle 2) \:=\:\frac{281^2 + 153^2 - 178^2}{2(281)(153)} \:=\:\frac{70,\!686}{85,\!986} \:=\:\frac{231}{281}$

Hence: . $\angle 1 = \angle 2 \quad\Rightarrow\quad CD \parallel AB \quad\Rightarrow\quad \angle BCD = 90^o$

Finally, the diagram looks like this:

Code:
    A o
* * 89
*   *  M
78 * - - o
*     : * 89
*     :   *  D
o - - : - - o
*     :     *
*     :     *
153 *     :     * 153
*     :     *
*     :     *
*     :     *
B o - - o - - o C
80  N  80
Draw segment $BM.$

We see that $MN\:=\:39 + 153 \:=\:192$

$BM$ is the hypotenuse of right triangle $M\!N\!B.$

. . Hence: . $B\!M^2 \:=\:80^2 + 192^2 \:=\:43,\!264$

Therefore: . $B\!M \:=\:208$