(i) Let us join Point D and B.

BD ⊥AC, DM ⊥ BC and DN ⊥ AB

Now from the figure we have,

DN || CB, DM || AB and ∠B = 90 °

Therefore, DMBN is a rectangle.

So, DN = MB and DM = NB

The given condition which we have to prove, is when D is the foot of the perpendicular drawn from B to AC.

∴ ∠CDB = 90° ⇒ ∠2 + ∠3 = 90° ……………………. (i)

In ∆CDM, ∠1 + ∠2 + ∠DMC = 180°

⇒ ∠1 + ∠2 = 90° ……………….. (ii)

In ∆DMB, ∠3 + ∠DMB + ∠4 = 180°

⇒ ∠3 + ∠4 = 90° ………………….. (iii)

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

∠1 = ∠3

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

∠2 = ∠4

In ∆DCM and ∆BDM,

∠1 = ∠3 (Already Proved)

∠2 = ∠4 (Already Proved)

∴ ∆DCM ∼ ∆BDM (AA similarity criterion)

BM/DM = DM/MC

DN/DM = DM/MC (BM = DN)

⇒ DM² = DN × MC

Hence, proved.

(ii) In right triangle DBN,

∠5 + ∠7 = 90° ……………….. (iv)

In right triangle DAN,

∠6 + ∠8 = 90° ………………… (v)

D is the point in triangle, which is foot of the perpendicular drawn from B to AC.

∴ ∠ADB = 90° ⇒ ∠5 + ∠6 = 90° ………….. (vi)

From equation (iv) and (vi), we get,

∠6 = ∠7

From equation (v) and (vi), we get,

∠8 = ∠5

In ∆DNA and ∆BND,

∠6 = ∠7 (Already proved)

∠8 = ∠5 (Already proved)

∴ ∆DNA ∼ ∆BND (AA similarity criterion)

AN/DN = DN/NB

⇒ DN² = AN × NB

⇒ DN² = AN × DM (Since, NB = DM)

Hence, proved.