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**Q. (1) In fig (10).9(2) It is given that AB = CD and AD = BC. Prove that ΔADC≅ΔCBA.**

**Solution:**

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

We have to prove ΔADC≅ΔCBA

Now,

Consider Δ ADC and Δ CBA.

We have

AB = CD [Given]

BC = AD [Given]

And AC = AC [Common side]

So, by SSS congruence criterion, we have

ΔADC≅ΔCBA

Hence proved

**Q. (2) In a Δ 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 Δ PQR, PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively

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 = ……(i) Using mid-point theorem,

We have

MN ∥ PQ and MN =

- MN = PL = LQ ……(ii)

Similarly, we have

LN ∥ QR and LN = (1/2)QR

- LN = QM = MR ……(iii)

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

PL = LQ = QM = MR = MN = LN

LN = MN

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