From the \(\triangle \)ABC, AB = AC
It is an isosceles triangle
Angles opposite to equal sides are equal
∴ \(\angle \)B =\(\angle \)C = 65^o
∴ \(\angle \)B +\(\angle \)C = 65^o + 65^o
= 130^o
We know that sum of three angles is a triangle = 180^o
\(\angle \)A+\(\angle \)B+\(\angle \)C = 180^o
\(\angle \)A + 130^o = 180^o
\(\angle \)A  = 180^o -130^o
\(\angle \)A  = 50^o
From \(\triangle \)DEF, \(\angle \)D = 50^o
∴ Sum of Remaining angles 180^o - 50^o = 130^o
DE = FD
∴ \(\angle \)D = \(\angle \)F
From \(\triangle \)ABC and \(\triangle \)DEF∴ \(\angle \)E = \(\frac { 130 }{ 2 } \)= 65^o
\(\angle \)A = \(\angle \)D = 50^o
\(\angle \)B =\(\angle \)E = 65^o
\(\angle \)C =\(\angle \)F = 65^o
∴ By AAA criteria \(\triangle \)DEF ~ \(\triangle \)ABC

  \(\frac { GU }{ BO} \)= \(\frac { 32 }{ 8 } \) = \(\frac { 4 }{ 1 } \)
\(\frac { UM }{ OX} \)= \(\frac { 48 }{ 12 } \)= \(\frac { 4 }{ 1 } \)
\(\frac { GM }{ BX} \)= \(\frac {52 }{ 13 } \)=\(\frac { 4 }{ 1 } \)
We find that \(\frac { GU }{ BO} \)= \(\frac { UM }{ OX} \)= \(\frac { GM }{ BX} \)= \(\frac { 4 }{ 1 } \)
That is their corresponding sides are proportional.
:. By SSS similarity \(\triangle \)GUM ~ \(\triangle \)BOX.

\(Given \triangle \mathrm{EAT} \sim \triangle \mathrm{BUN} \)
\(\therefore \angle \mathrm{E}=\angle \mathrm{B}, \angle \mathrm{A}=\angle \mathrm{U}, \angle \mathrm{T}=\angle \mathrm{N}\)
Sum of the angles of a triangle is 180^o
\(\angle \mathrm{E}+\angle \mathrm{A}+\angle \mathrm{T}=180^{\circ}\)
x + 2x + x + 40 = 180^o
4x = 180^o - 40^o = 140^o
\( x =\frac{140^{\circ}}{4}=35^{\circ} \)
\(x\therefore \angle \mathrm{E} =\angle \mathrm{B}=x=35^{\circ} \)
\(\angle \mathrm{A} =\angle \mathrm{U}=2 x \)
\(=2 \times 35^{\circ}=70^{\circ} \)
\(\angle \mathrm{T} =\angle \mathrm{N}=x+40^{\circ} \)
\(=35^{\circ}+40^{\circ}=75^{\circ} \)

\(\text { In } \triangle \mathrm{CUB} \text { and } \Delta \mathrm{CAT}\)
CU = CB (given)
\(\angle \mathrm{CUB}=\angle \mathrm{CBU}\)
\( \mathrm{UB} \| \mathrm{AT} \)
\(\angle \mathrm{CUB} =\angle \mathrm{CAT} \)
\(\angle \mathrm{CBU} =\angle \mathrm{CTA} \)
\(\angle \mathrm{C} =\angle \mathrm{C} \)
\(\therefore \Delta \mathrm{CUB} \sim \Delta \mathrm{CAT} \)
Also \(\triangle\) CAT is isosceles.
Hence proved.