# In the figure, D is a point on hypotenuse AC of $$\triangle$$ ABC, BD $$\perp$$ AC, DM $$\perp$$ BC and DN $$\perp$$ AB. Prove that (i) $${DM}^2$$ = DN.MC (ii) $${DN}^2$$ = DM.AN

## Solution :

We have : AB $$\perp$$ BC and DM $$\perp$$ BC.

So,        AB || DM

Similarly, we have :

BC $$\perp$$ AB and DN $$\perp$$ AB.

So,          CB || DN

Hence, quadrilateral BMDN is a rectangle.

$$\therefore$$  BM = DN

(i)  In triangle BMD, we have :

$$\angle$$1 + $$\angle$$ BMD + $$\angle$$2 = 180

or  $$\angle$$1 + 90 + $$\angle$$2 = 180

$$\implies$$  $$\angle$$1 + $$\angle$$2 = 90

Similarly, in triangle BMC, we have :

$$\angle$$3 + $$\angle$$4 = 90

Since BD $$\perp$$ AC, therefore

$$\angle$$2 + $$\angle$$3 = 90

Now, $$\angle$$1 + $$\angle$$2 = 90   and  $$\angle$$2 + $$\angle$$3 = 90

$$\therefore$$  $$\angle$$1 + $$\angle$$2 = $$\angle$$2 + $$\angle$$3

So,  $$\angle$$1 = $$\angle$$3

Also,  $$\angle$$3 + $$\angle$$4 = $$\angle$$2 + $$\angle$$3  $$\implies$$  $$\angle$$2 = $$\angle$$4

Thus, in triangles BMD and BMC, we have :

$$\angle$$1 = $$\angle$$3   and  $$\angle$$2 = $$\angle$$4

$$\therefore$$ By AA similarity, we have :

$$\triangle$$ BMD ~ $$\triangle$$ DMC

So,  $$BM\over DM$$ = $$MD\over MC$$

Since, BM = ND,

$$\implies$$  $$DN\over DM$$ = $$DM\over MC$$

So, $${DM}^2$$ = DN.MC

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

$$\triangle$$ BND ~ $$\triangle$$ AND

So,  $$BN\over DN$$ = $$ND\over NA$$

Since, BN = DM,

$$\implies$$  $$DM\over DN$$ = $$DN\over AN$$

So, $${DN}^2$$ = DN.MC