Sum of lengths of any two sides of a triangle is always greater than the third side. Hence, the given statement is false.

The sum of all the three angles of a triangle is ${180}^{\circ }$.

We know that obtuse angle is an angle which is greater than ${90}^{\circ }$ and less than ${180}^{\circ }$
Hence a triangle cannot have two obtuse angles.
 

The sum of 2 obtuse angles will be greater than (90 + 90), i.e. greater than ${180}^{\circ }$.

Since the sum of 2 angles of the triangle is more than ${180}^{\circ }$, the sum of three angles will be more than ${180}^{\circ }$ for sure.

This is not possible as the sum of 3 angles of a triangle is fixed i.e. ${180}^{\circ }$ and  cannot exceed this limit.

Thus, a triangle cannot have 2 obtuse angles.

Given $\angle$C = 60^o and BC = AC,
Therefore $\angle A=\angle B$  (base angles are equal in an isosceles triangle)
$\therefore$  by angle sum property of triangle, $\angle A=\angle B={60}^o$

We know that,  $\angle$x +$\angle$y = $\angle$A +$\angle$B  (Exterior angle property)
$\therefore \angle A+\angle B+\angle x+\angle y={60}^{\circ }+{60}^{\circ }+{120}^{\circ }={240}^{\circ }$

A triangle with two right angles is not possible, as the sum of all the three angles of a triangle is ${180}^{\circ }$.

$\angle$C = ${60}^{\circ }$​ and BC = AC,
By angle sum property of triangle, $\angle$A = $\angle$B = ${60}^{\circ }$​

Also,
$\angle ACB+\angle x+\angle y={180}^0$  [ BCD is straight line]
$⟹{60}^{\circ }+2y+y={180}^{\circ }$​ 
$⟹{60}^{\circ }+3y={180}^{\circ }$
$⟹3y={180}^{\circ }-{60}^{\circ }$​
$⟹y={40}^{\circ }$

Hence, $\angle B+y={60}^{\circ }+{40}^{\circ }={100}^{\circ }$​

      $\angle ACB+\angle ABC=\angle DAC$
                  [Exterior angle property]
     $\angle ACB+{60}^{\circ }={150}^{\circ }\angle ACB={90}^{\circ }\therefore \text{Reflex}\angle ACB={360}^{\circ }-{90}^{\circ }={270}^{\circ }=3\times {90}^{\circ }=3\text{right angles}$

We know that sum of the three angles of a triangle = ${180}^{\circ }$​

Therefore, $x+2x+3x={180}^{\circ }$​

$or,6x={180}^{\circ }$,
$x=\frac{180}{6}={30}^{\circ }$

A triangle can't have all the three angles greater than ${60}^{\circ }$,​ because the sum of all the three angles of a triangle is ${180}^{\circ }$.

Sum of lengths of any two sides of a triangle is always greater than the third side.
Hence,
                  AB + BM $>$ AM
                  AC + CM $>$ AM
Thus, AB + BC + CA $>$ 2AM

By Exterior Angle Property,
$\angle$ACD  = $\angle$A​ + $\angle$B
$\angle$B = $\angle$ACD - $\angle$A​
       = ${120}^{\circ }$​ - ${70}^{\circ }$
$\angle$B = ${50}^{\circ }$​