ABC

A

AD

BE

CF

E'

F'

E

F

BC

2E'F'=2AD+BC

A

135^{\circ}

BC=a

AC=b

AB=c

\angle A=\alpha

\angle B=\beta

\angle C=\gamma

AD=AC\sin\gamma=b\sin\gamma,

CF'=BC-BF'=a-BF\cos\beta=a-BC\cos\beta\cdot\cos\beta=

=a-a\cos^{2}\beta=a\sin^{2}\beta.

BE'=a-a\sin^{2}\gamma.

E'F'=BC-CF'-BE'=2a-a\sin^{2}\beta-a\sin^{2}\gamma.

2(2a-a\sin^{2}\beta-a\sin^{2}\gamma)=2b\sin\gamma+a,

\frac{b}{a}=\frac{\sin\beta}{\sin\alpha}

\cos(\beta+\gamma)=-\sin\alpha

\sin\alpha(1-2\sin^{2}\beta-2\sin^{2}\gamma=2\sin\beta\sin\gamma~\Leftrightarrow~

~\Leftrightarrow~\sin\alpha(1-(1-\cos2\beta+1-2\sin2\gamma)=\cos(\beta-\gamma)-\cos(\beta+\gamma)~\Leftrightarrow~

~\Leftrightarrow~\sin\alpha(2\cos(\beta-\gamma)\cos(\beta+\gamma)-1)=\cos(\beta-\gamma)+\cos\alpha~\Leftrightarrow~

~\Leftrightarrow~-\cos(\beta-\gamma)(1+2\sin\alpha\cos\alpha)=\cos\alpha+\sin\alpha~\Leftrightarrow~

~\Leftrightarrow~(\cos\alpha+\sin\alpha)(1+(\cos\alpha+\sin\alpha)\cos(\beta-\gamma))=0.

\alpha\gt90^{\circ}

\alpha\geqslant135^{\circ}

\alpha-45^{\circ}\geqslant90^{\circ}

1+(\cos\alpha+\sin\alpha)\cos(\beta-\gamma)=1+\sqrt{2}\cos(\alpha-45^{\circ})\cos(\beta-\gamma)\gt0,

-1\lt\sqrt{2}\cos(\alpha-45^{\circ})\cos(\beta-\gamma)\leqslant0

\cos(\beta-\gamma)\gt0

0^{\circ}\lt\alpha\lt135^{\circ}

\cos(\alpha-45^{\circ})\gt-\frac{1}{\sqrt{2}}

1+(\cos\alpha+\sin\alpha)\cos(\beta-\gamma)=1+\sqrt{2}\cos(\alpha-45^{\circ})\cos(\beta-\gamma)\gt

\gt1+\sqrt{2}\cdot\left(-\frac{1}{\sqrt{2}}\right)=0.

\cos\alpha+\sin\alpha=0,

\alpha=135^{\circ}