Răspuns: [tex]\bf \angle{BCD}=135^{\circ}[/tex]
Explicație pas cu pas:
Salutare!
[tex]\bf In~\triangle{BAC}-isoscel~avem:[/tex]
[tex]\bf \angle{BAC}=40^{\circ}[/tex]
[tex]\bf \angle{ABC}=\angle{ACB}[/tex]
[tex]\bf [AB]=[AC][/tex]
[tex]\bf \implies \angle{ABC}+\angle{BAC}+\angle{ACB}=180^{\circ}[/tex]
[tex]\bf 2\cdot \angle{ABC}+40^{\circ}=180^{\circ}[/tex]
[tex]\bf 2\cdot \angle{ABC}=180^{\circ}-40^{\circ}[/tex]
[tex]\bf 2\cdot \angle{ABC}=140^{\circ}~~~\Big|:2[/tex]
[tex]\boxed{\bf \angle{ABC}=70^{\circ}\implies \angle{ACB}=70^{\circ}}[/tex]
[tex]\bf In~\triangle{ACD}-isoscel~avem:[/tex]
[tex]\bf \angle{CAD}=50^{\circ}[/tex]
[tex]\bf \angle{ADC}=\angle{ACD}[/tex]
[tex]\bf [AD]=[AC][/tex]
[tex]\bf \implies \angle{ADC}+\angle{ACD}+\angle{CAD}=180^{\circ}[/tex]
[tex]\bf 2\cdot \angle{ACD}+50^{\circ}=180^{\circ}[/tex]
[tex]\bf 2\cdot \angle{ACD}=180^{\circ}-50^{\circ}[/tex]
[tex]\bf 2\cdot \angle{ACD}=130^{\circ}~~~\Big|:2[/tex]
[tex]\boxed{\bf \angle{ACD}=65^{\circ}\implies \angle{ADC}=65^{\circ}}[/tex]
[tex]\bf \angle{BCD}=70^{\circ}+65^{\circ}[/tex]
[tex]\boxed{\boxed{\bf \angle{BCD}=135^{\circ}}}[/tex]
Varianta b)
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