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D is a point on hypotenuse AC of triangle ABC, such that BD is perpendicular to AC, DM is perpendicular to BC and DN is perpendicular to AB. Prove that:1 DM2=DN ×MC 2 

DN2=DM ×AN

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(i) 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° … (1)
In ΔCDM,
∠1 + ∠2 + ∠DMC = 180°
⇒ ∠1 + ∠2 = 90° … (2)
In ΔDMB,
∠3 + ∠DMB + ∠4 = 180°
⇒ ∠3 + ∠4 = 90° … (3)
From equation (1) and (2), we obtain
∠1 = ∠3
From equation (1) and (3), we obtain
∠2 = ∠4
In ΔDCM and ΔBDM,
∠1 = ∠3 (Proved above)
∠2 = ∠4 (Proved above)
∴ ΔDCM ∼ ΔBDM (AA similarity criterion)

⇒ DM2 = DN × MC
(ii) In right triangle DBN,
∠5 + ∠7 = 90° … (4)
In right triangle DAN,
∠6 + ∠8 = 90° … (5)
D is the foot of the perpendicular drawn from B to AC.
∴ ∠ADB = 90°
⇒ ∠5 + ∠6 = 90° … (6)
From equation (4) and (6), we obtain
∠6 = ∠7
From equation (5) and (6), we obtain
∠8 = ∠5
In ΔDNA and ΔBND,
∠6 = ∠7 (Proved above)
∠8 = ∠5 (Proved above)
∴ ΔDNA ∼ ΔBND (AA similarity criterion)


⇒ DN2 = AN × NB
⇒ DN2 = AN × DM (As NB = DM)
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