Răspuns:
ΔABC isoscel ( AB=AC)
[BD bis. ===> AB/BC = AD/DC => BC=AB*DC/AD
[BD bis. ∡B => m(∡ABD)=m(∡DBC)=m(∡B)/2=72°/2 = 36°
in ΔBDC: m(∡BDC)= 180°- ( m(∡DBC)+ m(∡C) )
m(∡BDC)= 180°- 108°= 72° = m(∡C) ===> ΔBDC isoscel cu baza [DC] => BD=BC
m(∡ADB)+m(∡BDC)= 180° ( D∈(AC) ) => m(∡ADB)= 108°
in ΔADB: m(∡BAD)= 180° - ( m(∡ABD) + m(∡ADB) )
m(∡BAD)= 180°- 144°= 36° =m (∡ABD) => ΔADB isoscel cu baza [ AB] => AD=BD
avem rel. BC= AB* (DC/AD) | *AD => BC*AD= AB*DC
AB=AC
BD=BC iar AD=BD =>> BC=AD
Deci, BC*BC= AC*DC <=> BC²=AC*DC
