In $△ABO$ and $△CBO$

(i) $\angle BAO=\angle CAO={90}^{\circ }$ ........ (given)

(ii) BO = BO .........(common side)

(iii) AB = BC........ (given)

$\Rightarrow △ABO\cong △CBO$    ........ (RHS Postulate)
$\Rightarrow$ OA = OC.......(cpct)
$\Rightarrow$ $\angle ABO=\angle CBO={60}^{\circ }$ ........(cpct)

Consider $△ABC$ and $△ADC$

(i) AB = CD ...... (given)
(ii) BC = DA ...... (given)
(iii) AC = AC ...... (common)

$\therefore △ABC\cong △CDA$ ... (SSS Postulate)

$\Rightarrow \angle ABC=\angle CDA$ ..... (CPCT)

$\therefore x={48}^{\circ }$

 $\Rightarrow \angle BCA=\angle DAC$ ......(CPCT)

$\therefore y={56}^{\circ }$

In $△ABD$ and $△ACD$

(i) AB = AC .........(given)

(ii) BD = CD .........(given)

(iii) AD = AD ..........(common)

(iv)$△ABD\cong △ACD$ ......(SSS Postulate)

(v) $\angle BAD=\angle CAD$ .....(cpct)

$\therefore$ AD bisects $\angle$ BAC

(vi) $\angle ABD=\angle ACD$ .....(cpct)

If $\angle BAC={80}^{\circ }$, then $\angle ABD=\angle ACD={50}^{\circ }$ ( not ${80}^{\circ }$)

In the given figure:

In  $\Delta ABCand\Delta PQR$

AB = PR

BC = PQ

AC = QR

Therefore, $\Delta ABC\cong \Delta PQR$ by SSS criterion.
Hence, the statement is false.

In the given figure:

In  $\Delta ABCand\Delta PQR$

AB = PR

BC = PQ

AC = QR

Therefore, $\Delta ABC\cong \Delta RPQ$ by SSS criterion.

Since all congruent triangles are similar triangles. $\Delta ABCand\Delta RPQ$  are also similar.

In the given figure:

In  $\Delta ABCand\Delta PQR$

AB = PR (Given)

BC = PQ (Given)

AC = QR (Given)

Therefore, $\Delta ABC\cong \Delta PQR$ by SSS condition.

Therefore,

$\angle A=\angle R={50}^{\circ }$;

$\angle B=\angle P$;

$\angle C=\angle Q={60}^{\circ }$.

Using angle sum property in  $\Delta$ ABC,
$\angle A+\angle B+\angle C={180}^{\circ }$
$\Rightarrow$   ${50}^{\circ }+\angle B+{60}^{\circ }={180}^{\circ }$
$\Rightarrow$  $\angle B={70}^{\circ }$

In the given figure:

In  $\Delta ABCand\Delta PQR$

AB = PR = 5 cm

BC = PQ

AC = QR = 7 cm

Therefore, $\Delta ABC\cong \Delta PQR$ by SSS criterion.

Since perimeter = 18cm

$\Rightarrow$ AB + AC + BC = 18 cm,

$\Rightarrow$ BC = 18 - 5 - 7 = 6 cm.
Therefore, 
AB = PR = 5 cm

BC = PQ = 6 cm

AC = QR = 7 cm 
So, the answer is 7 cm.

In $\Delta ABD\text{and}\Delta BAC$,
AD = BC  (Given)
$\angle BAD$ = $\angle CBA$ (Given)
AB = BA  (Side common to both triangles)
Hence $\Delta ABD\cong \Delta BAC$
(by SAS congruence condition). 
Thus, $\angle ABD$ = $\angle BAC$
(congruent parts of congruent triangles). 

We know that if two triangles are congruent, then their corresponding parts are equal.
Since $\Delta DEF\cong \Delta BCA$, therefore $\angle E=\angle C$ and $FD=AB$.

In $\Delta ABCand\Delta PQR$​

AB = PQ = 4cm ,  (Given)

BC = QR = 5 cm,  (Given) 

AC = PR = 6 cm;  (Given)

Hence, $\Delta ABC\cong \Delta PQR$  (By SSS criterion).