Помогите с геометрией :)

3)<DBC= <ABC/2 =(90° -<BAC)/2 =(90° -36°)/2 =54°/2 =27°.<BDC =90 -<DBC =90° -27° =63°.----------------------------------------4) AB =BC ⇒ <A=<C .<A+<С =100°⇒<B =180° -100° =80°.<A+<A =100 °;<A =<C =50.внешние углы :<A₁ =180° -<A =180° -50° =130° ;  или <A₁ = <B +<C =80°+50° =130° .<С₁= 180° -<С =180° -50° =130°;   или <С₁= <B+ <A =80°+50° =130° .<B₁=180° -<B  =180° -80°= 100°.   или <B₁ =<A+<C =50°+50°  =100°.