Can you find the perimeter?

Label the diagram as follows:

By the exterior angle theorem, $\angle CDE = \alpha + \beta$ , and since base angles of isosceles triangle are congruent, $\angle CED = \angle CDE = \alpha + \beta$ , and by the exterior angle theorem, $\angle CBA = \beta$ . Therefore, $\triangle CDA \sim \triangle BEA$ by AA similarity, so that $\cfrac{w}{x} = \cfrac{w + z}{6}$ .

By the angle bisector theorem, $\cfrac{x}{25}$ = $\cfrac{6}{y}$ , which rearranges to $y = \cfrac{150}{x}$ , and $\cfrac{25}{w} = \cfrac{x}{z}$ , which rearranges to $w = \cfrac{25z}{x}$ .

Substituting $y = \cfrac{150}{x}$ and $w = \cfrac{25z}{x}$ into $\cfrac{w}{x} = \cfrac{w + z}{6}$ leads to $x^2 + 25x - 150 = 0$ , which has a solution of $x = 5$ for $x > 0$ .

If $x = 5$ , then $y = \cfrac{150}{x} = \cfrac{150}{5} = 30$ , so that the perimeter is $P = 25 + x + 6 + y = 25 + 5 + 6 + 30 = \boxed{66}$ .

Construction: Draw line segment $DM$ such that $DM\parallel BC$ and $M$ lies on $AB$ .

Let $\angle EAC=\angle BAE=\alpha$ and $\angle ACD=\angle DCE=\beta$ , by exterior angle property , $\begin{aligned} \angle CDE&=\angle CED=\angle DAC+\angle DCA=\alpha+\beta \\ \angle ABE&=\angle DMA=\angle CED-\angle EAB=\beta \end{aligned}$ Let $CD=CE=x$ , since $\angle CBM=\angle BCD=\beta$ and $DM\parallel BC,\;$ $BCDM$ is an isosceles trapezium $\implies BM=CD=x$ .

By Angle Bisector Theorem , $\frac{AC}{AB}=\frac{CE}{EB}\implies \frac{25}{AB}=\frac{x}{6} \implies AB=\frac{150}{x}$ Observe, $\angle DMA=\angle DCA=\beta,\; \angle DAC=\angle DAB=\alpha\implies \triangle ADM\cong \triangle ADC$ . So, we have, $\begin{aligned} AM&=AC \\ \frac{150}{x}-x&=25 \\ x^2+25x-150&=0 \\ (x-5)(x+30)&=0 \\ \implies x&=5 \end{aligned}$ Therefore, perimeter of $\triangle ABC=AB+BC+CA=30+11+25=\boxed{66}$