# Computing the distance

• Aug 10th 2012, 01:46 PM
Mhmh96
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
• Aug 10th 2012, 05:31 PM
Soroban
Re: Computing the distance
Hello, Mhmh96!

I assume you made a sketch . . .

Quote:

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