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.triangle

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

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