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In Fig. 6.57, D is a point on hypotenuse AC of $$\triangle$$ ABC, such that BD $$\perp$$ AC, DM $$\perp$$ BC and DN $$\perp$$ AB. Prove that: (i) $$DM^2 = DN . MC$$ (ii) $$DN^2 = DM . AN.$$

Let us join Point D and B.
(i) Given,
BD $$\perp$$ AC, DM $$\perp$$ BC and DN $$\perp$$ AB

Now from the figure we have,
DN || CB, DM || AB and $$\angle$$ 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.

$$\angle$$ CDB = 90°
$$\Rightarrow$$ $$\angle$$ 2 + $$\angle$$ 3 = 90° ……………………. (i)

In $$\triangle$$ CDM, $$\angle$$ 1 + $$\angle$$ 2 + $$\angle$$ DMC = 180°
$$\angle$$ 1 + $$\angle$$ 2 = 90° …………………………………….. (ii)

In $$\triangle$$ DMB, $$\angle$$ 3 + $$\angle$$ DMB + $$\angle$$ 4 = 180°
$$\angle$$ 3 + $$\angle$$ 4 = 90° …………………………………….. (iii)

From equation (i) and (ii), we get
$$\angle$$ 1 = $$\angle$$ 3

From equation (i) and (iii), we get
$$\angle$$ 2 = $$\angle$$ 4

In $$\triangle$$ DCM and $$\triangle$$ BDM,
$$\angle$$ 1 = $$\angle$$ 3 (Already Proved)
$$\angle$$ 2 = $$\angle$$ 4 (Already Proved)
$$\triangle$$ DCM ~ $$\triangle$$ BDM (AA similarity criterion)
$$\frac{BM}{DM} = \frac{DM}{MC}$$
$$\frac{DN}{DM} = \frac{DM}{MC}$$ (BM = DN)
$$DM^2 = DN × MC$$

Hence, proved.

(ii) In right triangle DBN,
$$\angle$$ 5 + $$\angle$$ 7 = 90° ……………….. (iv)

In right triangle DAN,
$$\angle$$ 6 + $$\angle$$ 8 = 90° ………………… (v)

D is the point in triangle, which is foot of the perpendicular drawn from B to AC.
$$\angle$$ ADB = 90° => $$\angle$$ 5 + $$\angle$$ 6 = 90° ………….. (vi)

From equation (iv) and (vi), we get,
$$\angle$$ 6 = $$\angle$$ 7

From equation (v) and (vi), we get,
$$\angle$$ 8 = $$\angle$$ 5

In $$\triangle$$ DNA and $$\triangle$$ BND,
$$\angle$$ 6 = $$\angle$$ 7 (Already proved)
$$\angle$$ 8 = $$\angle$$ 5 (Already proved)
$$\triangle$$ DNA ~ $$\triangle$$ BND (AA similarity criterion)
$$\frac{AN}{DN} = \frac{DN}{NB}$$
$$DN^2 = AN × NB$$
$$DN^2 = AN × DM$$ (Since, NB = DM)
Hence, proved.