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(i) \(DM^2 = DN . MC\)

(ii) \(DN^2 = DM . AN.\)

Answer :

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.

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