VEZI DESEN
Prin O ducem paralela MN ,la baze (N pe BC si M pe AD).
Triunghiurile : OCN
si ACB sunt asemenea (ON//AB) si avem ;
ON/AB=CN/CB.
Triunghiurile ;
OBN si DCB sunt asemenea (ON//CD) si avem
ON/CD=BN/BC.
adunamcele 2 relatii de mai sus si avem ;
ON/AB+ON/CD =CN/CB+BN/BC=(CN+BN)/BC=1
ON/8+ON/6=1 aducem la acelasi numitor:
6ON+8ON=48
14ON=48
ON=48/14=24/7
In ΔCAB avem:
CO/CA=ON/AB
CO/21=24/(7*8)
CO=(21*24)/(7*8)=3*3=9
AO=AC=CO=21-9=12cm
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VEZI DESEN
in ΔDBC
DN/NB=DQ/QC=2
=> DQ=CD*2/3=10
NQ/BC=DQ/DC
NQ/9=10/15=2/3
NQ=9*2/3=6
Aria ΔDNQ=NQ*DQ/2=6*10/2=30cm²
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VEZI DESEN
din teorema inaltimii:
MQ²=PQ*NQ=18*8=144
MQ=12
