\(\text{Using Pythagoras in}\ \Delta ABC: \)

\( AB=\sqrt{13^2-12^2}=\sqrt{25}=5 \)

\(\text{Scale factor}\ = \dfrac{FD}{AC} = \dfrac{39}{13} = 3 \)

\(\therefore ED = 3 \times AB = 3 \times 5 = 15 \)

\(\angle ACD \ \text{is common to}\ \Delta ACE\ \text{and}\ \Delta BCD \)

\(\angle CAE=\angle CBD=90^\circ \ \text{(Right cone)} \)

\(\therefore \Delta ACE\ \text{|||}\ \Delta BCD\ \ \text{(equiangular)}\)

\(\angle BDC = 180-(88+44) = 48^{\circ} \ \text{(angle sum of Δ)} \)

\(\angle CAD = \angle CDB = 48^{\circ} \)

\(\angle ACD=44^{\circ} \ \text{is common to}\ \Delta ACD\ \text{and}\ \Delta BCD \)

\(\therefore \Delta ACD\ \text{|||}\ \Delta DCB\ \ \text{(equiangular)}\)

\(\angle PQT = \angle SQR \ \text{(vertically opposite)} \)

\(\dfrac{PQ}{QS} = \dfrac{2.5}{5} = \dfrac{1}{2} \)

\(\dfrac{TQ}{QR} = \dfrac{2}{4} = \dfrac{1}{2} \)

\(\therefore \Delta PQT\ \text{|||}\ \Delta SQR\ \ \text{(sides adjacent to equal angles in proportion)}\)

\(\angle ACD = \angle BEA \ \text{(given)} \)

\(\angle BAE\ \text{is common to}\ \Delta ACD\ \text{and}\ \Delta BEA \)

\(\therefore \Delta ACD\ \text{|||}\ \Delta BEA\ \ \text{(equiangular)}\)

\(\text{Find unknown side}\ (x)\ \text{of smaller triangle}\)

\(\text{Using Pythagoras:}\)

\(x=\sqrt{5^2-4^2} = \sqrt{9} = 3\)
 

\(\dfrac{AC}{DE} = \dfrac{5}{15} = \dfrac{1}{3} \)

\(\dfrac{BC}{EF} = \dfrac{3}{9} = \dfrac{1}{3} \)

\(\angle ABC = \angle DEF = 90^{\circ} \ \ \text{(given)} \)

\(\therefore \Delta ABC\ \text{|||}\ \Delta DEF\ \ \text{(hypotenuse and second side of right-angled triangle in proportion)}\)

\(\dfrac{AB}{GH} = \dfrac{11}{22} = \dfrac{1}{2} \)

\(\dfrac{BC}{FG} = \dfrac{3.5}{7} = \dfrac{1}{2} \)

\(\angle ABC = \angle FGH = 95^{\circ} \ \ \text{(given)} \)
 

\(\therefore \Delta ABC\ \text{|||}\ \Delta HGF\ \ \text{(sides adjacent to equal angles in proportion)}\)

\(\angle ABE = \angle CBD\ \ \text{(vertically opposite)} \)

\(\angle AEB = \angle BCD\ \ \text{(alternate,}\ AE \parallel CD \text{)} \)

\(\therefore \Delta ABE\ \text{|||}\ \Delta DBC\ \ \text{(equiangular)}\)

\(\dfrac{RP}{AC} = \dfrac{18}{12} = \dfrac{3}{2} \)

\(\dfrac{QR}{BA} = \dfrac{12}{8} = \dfrac{3}{2} \)

\(\dfrac{PQ}{CB} = \dfrac{9}{6} = \dfrac{3}{2} \)
 

\(\therefore \Delta PQR\ \text{|||}\ \Delta CBA\ \ \text{(three pairs of sides in proportion)}\)