## RD Sharma Solutions Class 9 Chapter 10 Ex 10.4

**(1) In fig (10).9(2) It is given that AB = CD and AD = BC. Prove that \(\Delta ADC \cong \Delta CBA\).**

**Solution:**

Given that in the figure AB = CD Â and AD = BC.

We have to prove \(\Delta ADC \cong \Delta CBA\)

Now,

Consider \(\Delta\)

We have

AB = CD Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Given]

BC = AD Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Given]

And AC = AC Â Â Â Â Â Â Â Â Â Â Â Â [Common side]

So, by SSS congruence criterion, we have

\(\Delta ADC \cong \Delta CBA\)

Hence proved

**(2) In a \(\Delta\) PQR. IF PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN. **

**Sol:** Given that in \(\Delta\)

We have to prove LN = MN.

Join L and M, M and N, N and L

We have PL = LQ, QM = MR and RN = NP

[Since, L, M and N are mid-points of Pp. QR and RP respectively]

And also PQ = QR

- PL = LQ = QM = MR = \(\frac{PQ}{2}\)
= \(\frac{QR}{2}\) â€¦â€¦(i) Using mid-point theorem,

We have

MN \(\parallel\)

- MN = PL = LQ â€¦â€¦(ii)

Similarly, we have

LN \(\parallel\)

- LN = QM = MR â€¦â€¦(iii)

From equation (i), (ii) and (iii), we have

PL = LQ = QM = MR = MN = LN

LN = MN