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In figure, It is given that AB = CD and AD = BC. Prove that ΔADC ≅ ΔCBA.
Given that in the figure AB = CD and AD = BC.
We have to prove ΔADC ≅ ΔCBA
Now,
Consider ΔADC and ΔCBA.
We have
AB = CD [Given]
BC = AD [Given]
And AC = AC [Common side]
So, by SSS congruence criterion, we have
ΔADC ≅ ΔCBA
Hence proved
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.
Given that in ΔPQR, 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 = PQ/2 = QR/2 ... (i) Using mid-point theorem,
MN ∥ PQ and MN = PQ/2
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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Chapter 10: Congruent Triangles Exercise –...