Q. 24.2( 103 Votes )

# In Fig. 6.57, D is a point on hypotenuse AC of Δ ABC, DM ⊥ BC and DN ⊥ AB. Prove that:

(i) (ii)

Answer :

(i) **To Prove: DM ^{2} = 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)

DM^{2} = DN × MC

**Hence, Proved.**

(ii)

To Prove: DN

^{2}= 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

∠ADB = 90°

∠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

DN^{2} = AN x NB

DN^{2}= AN x DM (As NB = DM)

Hence, Proved

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