In Fig. 6.57, D is a point on hypotenuse AC of Δ ABC, DM ⊥ BC and DN ⊥ AB. Prove that:

(i) (ii)

(i)Let us join DB

We have, DN || CB,

DM || AB,

And ∠B = 90°

DMBN is a rectangle

DN = MB and DM = NB

The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC

∠CDB = 90°

∠2 + ∠3 = 90° (i)

In ΔCDM,

∠1 + ∠2 + ∠DMC = 180°

∠1 + ∠2 = 90° (ii)

In ΔDMB,

∠3 + ∠DMB + ∠4 = 180°

⇒∠3 + ∠4 = 90° (iii)

From (i) and (ii), we get

∠1 = ∠3

From (i) and (iii), we get

∠2 = ∠4

In ΔDCM and ΔBDM,

∠1 = ∠3 (Proved above)

∠2 = ∠4 (Proved above)

ΔDCM similar to ΔBDM (AA similarity)

BM/DM = DM/MC

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

DM² = DN × MC

(ii) In right triangle DBN,

∠5 + ∠7 = 90° (iv)

In right triangle DAN,

∠6 + ∠8 = 90° (v)

D is the foot of the perpendicular drawn from B to AC

∠ADB = 90°

∠5 + ∠6 = 90° (vi)

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

∠6 = ∠7

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

∠8 = ∠5

In ΔDNA and ΔBND,

∠6 = ∠7 (Proved above)

∠8 = ∠5 (Proved above)

Hence,

ΔDNA similar to ΔBND (AA similarity criterion)

AN/DN = DN/NB

DN² = AN * NB

DN²= AN * DM (As NB = DM)