Question

14. In Figure 17, AC, AD and AE are joined. Prove that <FAB + <ABC + <BCD + <CDE + <DEF + <EFA = 720° = ​

Answer 1

Answer:

by angle addition property of triangle,

in triangle ABC

angle BAC+CBA+ACB=180°_(1)

SIMILARLY

ANGLES,

DAC+ACD+CDA=180°_(2)

EAD+EDA+DEA=180°_(3)

EAF+EFA+AEF=180°_(4)

FROM FIGURE

EAF+EAD+DAC+BAC=180°_(5)

BCA+ACD=BCD_(6)

ADC+ADE=CDE_(7)

AED+AEF=DEF_(8)

ADDING (1),(2),(3),(4) AND FROM (5),(6),(7),(8)

<FAB + <ABC + <BCD + <CDE + <DEF + <EFA = 720°