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

(i) DM^{2} = DN.MC (ii) DN^{2} = DM.AN

**Solution:**

ΔMBN is a rectangle.

DM = BN and DN = BM......................... (i)

In ΔDCM

∠DCM + ∠DMC + ∠CDM = 180°

∠DCM + 90° + ∠CDM = 180°

∠DCM + ∠CDM = 90°........................ (ii)

But- ∠CDM + ∠BDM = 90°................. (iii)

Since BD ⊥ AC

From (ii) and (iii)

∠DCM = ∠BDM................ (iv)

In ΔBDM

∠DBM + ∠BDM = 90°................ (v)

Since, DM ⊥ BC

From (iii) and (v)

∠CDM = ∠DBM.............(vi)

Now in ΔDCM and ΔDBM

ΔDCM ~ ΔBDM (From (iv) and (vi), AA criterion)

DM/BM = MC/DM (Corresponding sides are in same ratio)

DM² = BM.MC

DM² = DN.MC [from (i) DN = BM]

(iii) In ΔBDN

∠BDN + ∠DBN = 90° (Since DN ⊥ AB) (vii)

But ∠ADN + ∠BDN = 90° (Since BD ⊥ AC) (viii)

From (vii) and (viii)

∠DBN = ∠ADN................. (ix)

In ΔADN

∠DAN + ∠ADN = 90° (Since DN ⊥ AB) (x)

But ∠BDN + ∠ADN = 90° (Since BD ⊥ AC) (xi)

From (xi) and (x)

∠DAN = ∠BDN.................. (xii)

Now in

ΔBDN and ΔDAN ,

ΔBDN ~ ΔDAN (From (ix) and (xii), AA criterion)

BN/DN = DN/AN (Corresponding sides are in same ratio)

DN² = BN.AN

DN² = DM.AN [from (i) BN = DM ]

**Video Solution:**

## In Fig. 6.57, D is a point on hypotenuse AC of ΔABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that: (i) DM^{2} = DN.MC (ii) DN^{2} = DM.AN

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.6 Question 2:

In Fig. 6.57, D is a point on hypotenuse AC of ΔABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that: (i) DM^{2} = DN.MC (ii) DN^{2} = DM.AN

In the above figure, D is a point on hypotenuse AC of ΔABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Hence it is proved that (i) DM^{2} = DN.MC (ii) DN^{2} = DM.AN