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
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.
Draw segment AD.Code:A o 178 * * * D * * o * * * 231 * * * * 281 * * * * * 153 * * * * * * * ** B o * * * * * * * * * o C 160
We find that: .$\displaystyle 160^2 + 231^2 \:=\:281^2$
. . Hence: .$\displaystyle \angle B \,=\,90^o.$
Now the diagram looks like this:
Let $\displaystyle \angle 1 = \angle BAC,\;\angle2 = \angle ACD$Code:A o * ** * 1 * * 178 * * * * * * * * * D 231 * * o * 281 * * * * * * * 2* 153 * * * * ** B o * * * * * * * o C 160
Law of Cosines:
$\displaystyle \cos(\angle 1) \:=\:\frac{231^2 + 281^2 - 160^2}{2(231)(281)} \:=\:\frac{106,\!722}{129,\!822} \:=\:\frac{231}{281}$
$\displaystyle \cos(\angle 2) \:=\:\frac{281^2 + 153^2 - 178^2}{2(281)(153)} \:=\:\frac{70,\!686}{85,\!986} \:=\:\frac{231}{281}$
Hence: .$\displaystyle \angle 1 = \angle 2 \quad\Rightarrow\quad CD \parallel AB \quad\Rightarrow\quad \angle BCD = 90^o$
Finally, the diagram looks like this:
Draw segment $\displaystyle BM.$Code:A o * * 89 * * M 78 * - - o * : * 89 * : * D o - - : - - o * : * * : * 153 * : * 153 * : * * : * * : * B o - - o - - o C 80 N 80
We see that $\displaystyle MN\:=\:39 + 153 \:=\:192$
$\displaystyle BM$ is the hypotenuse of right triangle $\displaystyle M\!N\!B.$
. . Hence: .$\displaystyle B\!M^2 \:=\:80^2 + 192^2 \:=\:43,\!264$
Therefore: .$\displaystyle B\!M \:=\:208$