Solution:
According to question,

AB = AC,         ∠B = ∠C
∠A + ∠B + ∠C = 180°
∠A + 2∠B = 180°
∠A = 180° - 70°
∠A = 110°
Note : In isosceles triangle median bisects the opposite side and make angle 90° on opposite side. It also bisects the vertex angle.
∠BAD = $$\frac{{\angle A}}{2}$$
∠BAD = $$\frac{{{{110}^ \circ }}}{2}$$
∠BAD = 55°

Solution:

We know that,
AD² = CD × BD
AD² = 4 × 3
AD² = 12
AD = $$2\sqrt 3 $$  cm

Solution:
Median of right angle
$$\eqalign{ = \frac{{EF}}{2} \cr = \frac{{12}}{2} \cr = 6\,{\text{cm}} \cr} $$

Solution:
According to question,
Given :

   ∠BAD = 109°
   ∠ACB = 72°
∴ ∠ACD = 180° - 72°
   ∠ACD = 108°
∴      AC = CD
   ∠CAD = ∠CDA
In ΔCDA
∠CAD + ∠CDA + ∠DCA = 180°
2∠CAD + 108° = 180°
2∠CAD = 180° -108°
2∠CAD = 72°
∠CAD = $$\frac{{{{72}^ \circ }}}{2}$$
∠CAD = 36°
∴ ∠CAB = 109° - 36°
∠CAB = 73°
In ΔABC
∠ABC + ∠ACB + ∠CAB = 180°
∠ABC + 72° + 73° = 180°
∠ABC + 145° = 180°
∠ABC = 180° - 145°
∠ABC = 35°

Solution:
According to question,

Given :
∠ABC = 60°
∠BCA = 80°
∠BIC = ?
∠BAC = 40°
∴ ∠BIC = 90° + $$\frac{1}{2}$$ × 40°
∠BIC = 110°

Solution:
$$\eqalign{ \frac{{{\text{area}}{\mkern 1mu} \left( {\vartriangle ABC} \right)}}{{{\text{area}}{\mkern 1mu} \left( {\vartriangle PMR} \right)}} \cr = \frac{{{{\left( 7 \right)}^2}}}{{\frac{1}{2} \times {{\left( 4 \right)}^2}}} \cr = \frac{{49}}{8} \cr} $$

Solution:
∵ Sum of all angles of a triangle = 180°
So, ∠BAC = 180 - (90 + 70)
∠BAC = 20°

So, ∠BIC = 90 + $$\frac{1}{2}$$ ∠A
∠BIC = 90° + $$\frac{1}{2}$$ × 20°
∠BIC = 100°