AB² = AD² + BD² ------- (1)

In Right angled $\Delta$ ACD,

AC² = AD² + BD² ------- (2)

By (1) - (2)

AB² - AC² = BD² - CD²

AB² - BD² = AC² - CD².

A triangle in which one of the median is perpendicular to the side it meets, is necessarily an isosceles triangle.

So, We can say that $\Delta$ BCD and &delta ABC will be similar.

According to property of similarity,

AB/12 = 12/9

Hence,

AB = 16

AC/6 = 12/9

AC = 8

Hence, AD = 7 and AC = 8

Now,

Perimeter of Delta; ADC / Perimeter of &delta BDC,

= (6+7+8)/(9+6+12)

= 21/27 = 7/9.

Let BC = x and Ad = y, then as per bisector theorem,

BD /DC = AB /AC = 4 /3

Hence, BD = 4x/7 and DC = 3x/7

Now, in $\Delta$ ABD using cosine rule,

cos 30⁰ = [{4² +y² - (16x²/49)} /2*3*y]

Or, 4 $\sqrt{3y}$ =[{16 +y² - (16x²/49)}] --------- (i)

Similarly in $\Delta$ ADC,

cos 30⁰ = [{3² +y² - (9x²/49)} /2*3*y]

Or, 3 $\sqrt{3y}$ =[{9 +y² - (9x²/49)}] --------- (ii)

From equation (i) and (ii), we get

y = 12 $\sqrt{3 }$/7.

QR +2 = 2PQ

QR = 2PQ - 2 ------- (2)

PR = PQ + 10 ----- (2)

Perimeter = 40 m

PQ + QR + Rp = 40

Putting the value of PQ and QR from equation (1) and (2),

Pq + 2PQ - 2 + PQ +10 = 40

4PQ = 32

PQ = 8 cm which is the smallest side of the triangle.