Đặt $AF=x;AE=y$
Áp dụng định lí Cê-va vào $\Delta ABC$ ta có:
$\frac{AF}{FB}.\frac{DB}{DC}.\frac{EC}{EA}=1\Leftrightarrow \frac{x}{c-x}.\frac{c}{b}.\frac{b-y}{y}=1\Leftrightarrow \frac{cx}{by}.\frac{b-y}{c-x}=1\Leftrightarrow \frac{c^2x^2}{b^2y^2}.\frac{(b-y)^2}{(c-x)^2}=1 \,(1)$
Theo GT: $\frac{1}{AB^{2}}+\frac{1}{AE^{2}}=\frac{1}{AC^{2}}+\frac{1}{AF^{2}}\Leftrightarrow \frac{1}{AF^{2}}-\frac{1}{AB^{2}}=\frac{1}{AE^{2}}-\frac{1}{AC^{2}}\Leftrightarrow \frac{AB^2-AF^2}{AF^2AB^2}=\frac{AC^2-AE^2}{AC^2AE^2}\Leftrightarrow \frac{c^2-x^2}{x^2c^2}=\frac{b^2-y^2}{b^2y^2}\Leftrightarrow \frac{x^2c^2}{b^2y^2}=\frac{c^2-x^2}{b^2-y^2}\,(2)$
Từ (1) và (2) suy ra: $\frac{(b-y)^2}{(c-x)^2}.\frac{c^2-x^2}{b^2-y^2}=1$
$\Leftrightarrow \frac{b-y}{b+y}.\frac{c+x}{c-x}=1\Leftrightarrow \frac{b-y}{b+y}=\frac{c-x}{c+x}\Leftrightarrow \frac{b-y}{b+y}+1=\frac{c-x}{c+x}+1\Leftrightarrow \frac{b}{b+y}=\frac{c}{c+x}\Leftrightarrow \frac{y}{b}=\frac{x}{c}\Leftrightarrow \frac{x}{c-x}=\frac{y}{b-y}$
Mà: $\frac{x}{c-x}.\frac{c}{b}.\frac{b-y}{y}=1\Leftrightarrow\frac{y}{b-y}.\frac{c}{b}.\frac{b-y}{y}=1\Leftrightarrow b=c$
Vậy $\Delta ABC$ cân tại A