Răspuns:
Explicație pas cu pas:
a)CE⊥AB inΔCEB CE=AD=x EB=AB-CD=23-18=5
CE²=BC²-EB²=21²-5²=16×26 ⇒x=4√26
b) CE⊥AB
in ΔCEB z²=25²-20²=5×45 ⇒z=3x5=15 AE=28-15=13
AF⊥DC DF-15-13=2
in ΔADF y²=DF²+CE²=20²+2²=404 ⇒y=4√101
c) AF⊥CD DF=(DC=AB)/2=(32-12)/2=20/2=10
in ΔADF u²=AD²+DF²=26²+10²=676+100=776 u=2√194
d) CE⊥AB EB=(AB-CD)/2=(28-12)/2=16/2=8
in ΔCEB v²=CE²+EB²=20²+8²=400+64=464 v=4√29
