# In Fig. 6.57, D is a point on hypotenuse AC of Δ ABC, DM ⊥ BC and DN ⊥ AB. Prove that:(i) (ii)  (i) To Prove: DM2 = DN. MC

Construction: join DB

We have, DN || CB,

DM || AB,

And B = 90° (Given)

As opposite sides are parallel and equal and also each angle is 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°

Now from the figure we can say that

2 + 3 = 90°              ........eq(i)

In ΔCDM,

1 + 2 + DMC = 180°   [ Sum of angles of a triangle = 180°]

1 + 2 = 90°           ...........eq(ii)

In Δ DMB,

3 + DMB + 4 = 180°    [ Sum of angles of a triangle = 180°]

⇒∠3 + 4 = 90°          ........eq(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)

(AA Similarity : When you have two triangles where one is a smaller version of the other, you are looking at two similar triangles.)

BM/DM = DM/MC

Cross multiplying we get,

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

DM2 = DN × MC

Hence, Proved.

(ii)
To Prove: DN2= AN x DM

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

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)

( AA similarity Criterion: When you have two triangles where one is a smaller version of the other, you are looking at two similar triangles.)

AN/DN = DN/NB

DN2 = AN x NB

DN2= AN x DM (As NB = DM)

Hence, Proved

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