Răspuns :
AD≡ BD≡CD atunci Δ DAB≡ Δ DBC
⇒ Δ ABD isoscel de baza AB
m∡ABD = m∡DAB = α si m∡ADB =180° - m∡ABD - m∡DAB
m∡ADB= 180° - α - α =180° - 2α
Δ BDC isoscel de baza BC
m∡DBC = m∡DCB = α
si m∡BDC = 180° - m∡DBC- m∡DCB=
= 180° -α - α =180° -2α
m∡ADC =180° = m∡ADB + m∡CDB
180° =180° -2α + 180° -2α
4α = 360° -180°
4α =180° ⇒ α =180° :4 = 45°
m∡ABC= m∡ABD + m∡DBC = α + α =2α =2·45° =90°
⇒ Δ ABC cu m∡B =90°
⇒ Δ ABD isoscel de baza AB
m∡ABD = m∡DAB = α si m∡ADB =180° - m∡ABD - m∡DAB
m∡ADB= 180° - α - α =180° - 2α
Δ BDC isoscel de baza BC
m∡DBC = m∡DCB = α
si m∡BDC = 180° - m∡DBC- m∡DCB=
= 180° -α - α =180° -2α
m∡ADC =180° = m∡ADB + m∡CDB
180° =180° -2α + 180° -2α
4α = 360° -180°
4α =180° ⇒ α =180° :4 = 45°
m∡ABC= m∡ABD + m∡DBC = α + α =2α =2·45° =90°
⇒ Δ ABC cu m∡B =90°