Hello, Coach!
Triangle ABC is isosceles with $\displaystyle AC=BC=13,\;AB=10.$
$\displaystyle M \text{ and }N$ are the feet of the perpendiculars from $\displaystyle A \text{ and }B\text{ to }BC\text{ and }AC$, resp.
Find $\displaystyle BN.$ Code:
C
*
/|\
/ | \
/ | \
/ | \
13 / | \ 13
/ 12| \
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/ | \
* - - - - + - - - - *
A 5 D 5 B
We see that $\displaystyle AD = 5,\;AC = 13.$
. . From Pythagorus: .$\displaystyle CD = 12$
Code:
C
*
/|\
/ | \
/ | \
/ | \
N / | \
* | \
/ * | \
/ | * \
/ | * \
* - - - - + - - - - *
A D B
We have similar right triangles: .$\displaystyle \Delta BNA \sim \Delta CDA$
. . (They share $\displaystyle \angle A$.)
Code:
C
*
/|
/ |
/ |
/ |
N 13 / | 12
* / |
/ * / |
/ * / |
/ * / |
* - - - - - - - - - * * - - - - *
A 10 B A 5 D
Hence, we have: .$\displaystyle \frac{BN}{10} \:=\:\frac{12}{13} $
Therefore: .$\displaystyle \boxed{BN \:=\:\frac{120}{13}}$