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Trapezul ABCD cu ABIIDC are AD=DC,m(DCA)=α si m(ACB)=β.Aflati masurile unghiurilor trapezului.                           

Răspuns :

[tex]m\left(\widehat{C}\right)=\alpha+\beta[/tex]
[tex]m(\widehat{B}=180^{\circ}-m(\widehat{C})=180^{\circ}-\alpha-\beta[/tex]
Triunghiul ADC este isoscel, deci [tex]m(\widehat{DAC})=\alpha[/tex]
Atunci [tex]m(\widehat{D})=180^{\circ}-2\alpha[/tex]
[tex]m(\widehat{A})=180^{\circ}-m(\widehat{D})=2\alpha[/tex]