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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.