Triangles Test ...

In $$\Delta ABC$$ and $$\Delta DEF$$, AB = DF and $$\angle A = \angle D$$. The two triangles will be congruent by SAS axiom if :

In $$\Delta PQR, \angle P = 60^{\circ}$$ and $$\angle Q = 50^{\circ}$$. Which side of the triangle is the longest ?

In the given figure , which of the following statement is true ?

If $$\Delta ABC \cong \Delta DEF$$ by SSS congruence rule then :

In $$\Delta ABC, \angle B = 30^{\circ}, \angle C = 80^{\circ}$$ and $$\angle A = 70^{\circ}$$ then,

In triangles ABC and DEF, AB $$=$$ FD and $$\angle A = \angle D$$. The two triangles will be congruent by
SAS axiom if :

In $$\Delta PQR,$$ if $$\angle R > \angle Q,$$ then:

The construction of a triangle ABC, given that BC = 3 cm is possible when difference of AB and AC is equal to :

ABC is an isosceles triangle with AB $$= $$AC and D is a point on BC such that  $$AD \perp BC$$ (Fig. 7.13). To prove that $$\angle BAD = \angle CAD,$$ a student proceeded as follows:

$$\Delta ABD$$ and $$ \Delta ACD,$$
AB $$=$$ AC (Given)
$$\angle B = \angle C$$   (because AB $$=$$ AC)
and $$\angle ADB = \angle ADC$$
Therefore, $$\Delta ABD \cong \Delta ACD (AAS)$$
So, $$\angle  BAD = \angle CAD (CPCT)$$
What is the defect in the above arguments?

Given $$\Delta OAP \cong  \Delta OBP$$ in figure, the criteria by which the triangles are congruent is: