Question

Trapezul ABCD cu AB II DC are AD=DC, m (ung. DCA)=α⁰ si m(ung. ACB)=β⁰. Aflati masurile unghiurilor trapezului.

Answer 1

Trapezul este isoscel, deci are unghiurile alaturate unuei baze congruente.

$m(\widehat{ADC})=m(\widehat{DCB})=\alpha^{\circ}+\beta^{\circ}$

$m(\widehat{DAB})=m(\widehat{ABC})=180^{\circ}-(\alpha^{\circ}+\beta^{\circ})$

Answer 2

AD=DC=>m(<DAC)=m(<ACD)=alfa
m(<ADC)=180-2alfa
DC||AB=><DCA=<BAC=>m(<BAC)=alfa
m(<ABC)=360-2alfa-(180-2alfa)-alfa-beta=180alfa-beta
In concluzie:
m(A)=2alfa
m(<B)=180-alfa-beta
m(C)=alfa+beta
m(D)=180- 2alfa