Question

ABC is a right triangle, right angled at B and D is the foot of the perpendicular drawn from B on AC. If DM perpendicular to BC and DN perpendicular to AB, prove that DM² = DN × MC and DN² = DM × AN.

Answer 1

We have,

AB⊥BC and DM⊥BC

⇒ AB∣∣DM

Similarly, we have

CB⊥AB and DN⊥AB

⇒ CB∣∣DN

Hence, quadrilateral BMDN is a rectangle.

∴  BM=ND

(i) In △BMD, we have

∠1+∠BMD+∠2=1800

⇒ ∠1+900+∠2=1800

⇒ ∠1+∠2=900

Similarly, in △DMC, we have

∠3+∠4=900

Since BD⊥AC. Therefore,

∠2+∠3=900

Now, ∠1+∠2=900 and ∠2+∠3=900

⇒ ∠1+∠2=∠2+∠3

⇒ ∠1=∠3

Also, ∠3+∠4=900 and ∠2+∠3=900

⇒ ∠3+∠4=∠2+∠3⇒∠2=∠4

Thus, in △′sBMD and DMC, we have

∠1=∠3 and ∠2=∠4

So, by AA-criterion of similarity, we have

    △BMD∼△DMC

⇒ DMBM​=MCMD​

⇒ DMDN​=MCDM​             [∵BM=ND]

⇒ DM2=DN×MC        [Hence proved]

(ii) Proceeding as in (i), we can prove that 

     △BND∼△DNA

⇒ DNBN​=NAND​

⇒  DNDM​=ANDN​

⇒ DN2=DM×AN    [Hence proved]

Answer 2

We have,

AB⊥BC and DM⊥BC

⇒ AB ∣∣ DM

Similarly, we have

CB⊥AB and DN⊥AB

⇒ CB ∣∣ DN

Hence, quadrilateral BMDN is a rectangle.

∴  BM = ND

(i) In △BMD, we have

∠1 +∠BMD + ∠2 = 180°

⇒ ∠1 + 90° + ∠2 = 180°

⇒ ∠1 + ∠2 = 90°

Similarly, in △DMC, we have

∠3 + ∠4 = 90°

Since BD⊥AC. Therefore,

∠2 + ∠3 = 90°

Now, ∠1 + ∠2 = 90° and ∠2 + ∠3 = 90°

⇒ ∠1 + ∠2 = ∠2 + ∠3

⇒ ∠1 = ∠3

Also, ∠3 + ∠4 = 90° and ∠2 + ∠3 = 90°

⇒ ∠3 + ∠4 = ∠2 + ∠3 

⇒ ∠2 = ∠4

Thus, in △′s BMD and DMC, we have

∠1 = ∠3 and ∠2 = ∠4

So, by AA-criterion of similarity, we have

△BMD ~ △DMC

(ii) Proceeding as in (i), we can prove that