If two angles have the same measure, they are congruent. Also, if two angles are congruent, their measures are same.

Hence, the measure of the other angle is also ${60}^o$.

If all the corresponding angles of two triangles are equal, then triangles will have the same shape, but not necessarily the same size and hence may not be congruent. See the figure below. The corresponding angles of both triangles are equal, but still their sizes are not same and hence they are not congruent.

For two angles to be congruent, they should coincide when superimposed. This is only possible if both the angles are equal. Hence, the given statement is true.

Criteria for congruence of triangles are SSS, SAS, ASA, AAS, RHS.

SSS- Two  triangles are congruent if all the 3 corresponding sides of the given triangles are equal. 

SAS- Two triangles are congruent if 2 corresponding sides of the given triangles and the corresponding angle between those sides are equal to each other. 

ASA- Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding angles and sides of other triangles. 

AAS- Two triangles are congruent if two pairs of corresponding angles and a pair of opposite sides are equal.

RHS- If the hypotenuse and a side of a right angled triangle are congruent with the hypotenuse and the corresponding side
of the other right angled triangle, then the two triangles are congruent with each other.

Consider $\Delta BOC$  and $\Delta AOD$
1) AD = BC  ( Given ) 
2) $\angle CBO=\angle DAO$= 90°
3) $\angle BOC=\angle AOD$ ......(vertically opposite angles)
$\therefore \Delta BOC\cong \Delta AOD$ ....[AAS Criterion]

$\Rightarrow$ OC = OD ....(congruent parts of congruent triangle)

Since, $\Delta BOC\cong \Delta AOD$, then $\angle BOC={30}^o$. From angle sum property of triangle in $\Delta BOC$, the measure of $\angle BCO$ is ${60}^o$.

Since, $\Delta BOC\cong \Delta AOD$,

Corresponding angles of the triangles are equal. It gives,
$\angle BOC={30}^o$ 
$\angle BOC+\angle BCO+\angle OBC={180}^o$ [Angle sum property]  
$\angle BCO$​​​​​​​ = 180 - (90 + 30) = 180 - 120 = ${60}^o$

Given, $AB=AC$
$⟹94=6x+4⟹6x=90⟹x=15$
 

In $△ABD$ and $△ACD$,
(i) $\angle ADB=\angle ADC={90}^{\circ }$ ... (given)
(ii) AB = AC ... (given)
(iii) AD = AD ... (common side)
$\Rightarrow$$△ABD\cong △ADC$ ... (RHS congruence rule)
Then, $BD=CD$ ... (CPCT)
$\Rightarrow 2y–7=11\Rightarrow 2y=18\Rightarrow y=9$

If all the side lengths of one triangle are equal to the side lengths of another triangle, then the triangles are congruent. This is called SSS criterion.

In two triangles,

If a pair of corresponding angles and the included side are equal, then they are congruent [ASA congruence criterion].

If a pair of corresponding angles and a non-included side are equal, then they are congruent [AAS congruence criterion].

Therefore, given statement is true.

In $△ABD$ and $△ADC$

(i) $\angle ADB=\angle ADC={90}^{\circ }$ .......(given)

(ii) AD = AD ....... (common)

(iii) AB = AC ....... (given)

(iv) $△ABD\cong △ADC$....... (RHS Postulate)

(v) BD = DC …… (cpct)

(vi) $\angle ABD$ = $\angle ACD={60}^{\circ }$ .......(cpct)

Hence (A) and (C)