Ducem inaltimiile din A si B si notam AN si BM
Δ BCM-Δdrept (M =90 grade)
sin C=cateta opusa\ipotenuza
sinC=BM\BC
√2\2=BM\BC
√2\2=BM\4√2
BM=4√2*√2\2=8\2=4
BM=4 cm
Acum putem afla CM
CM²=BC²-BM²=4√2²-4²=32-16=16
CM=√16=6
Trapezul fiind isoscel CM=DN=4
Stiim ca DC=10
putem afla MN
MN=DC-CM-DN=10-4-4=2
MN=AB=2
Diagonalele in trapezul isoscel sunt egale
ΔBDM-Δdreptunghic (M=90 grade)
DM=DN+MN=4+2=6
BD²=BM²+DM²=4²+6²=16+36=52
BD=√52=2√13
BD=AC=2√13
