In trapezium \(LMNO\), where \(LM||NO\) and \(LO=MN\), show that:
(i) \(\angle L = \angle M\)
(ii) \(\angle N = \angle O\)
(iii) \(△LMN≅△MLO\)
(iv) Diagonals \(LN\) and \(MO\) are equal in length.
(i) \(\angle L = \angle M\)
(ii) \(\angle N = \angle O\)
(iii) \(△LMN≅△MLO\)
(iv) Diagonals \(LN\) and \(MO\) are equal in length.
Proof for (i): \(∠L = ∠M\)
\(LM || ON\) , \(LE || ON\) Also \(LO || NE\) then,
\(LENO\) is a parallelogram.
Therefore, \(LO = NE\) -----(1) [Opposite sides of the parallelogram are equal]
But \(LO = MN\) [Given] ----(2)
From (1) and (2), \(MN = NE\)
Now, in \(∆MNE\), we have
\(MN = NE\)
\(LENO\) is a parallelogram.
Therefore, \(LO = NE\) -----(1) [Opposite sides of the parallelogram are equal]
But \(LO = MN\) [Given] ----(2)
From (1) and (2), \(MN = NE\)
Now, in \(∆MNE\), we have
\(MN = NE\)
\(∠NEM = ∠NME\) ----(3)
Also, \(∠LMN + ∠NME = 180^°\) ----(4) [] and
\(∠L + ∠NEM = 180^°\) ----(5) [Co-interior angles of a parallelogram LENO]
From (4) and (5), we get
\(∠LMN + ∠NME = ∠L + ∠NEM\)
\(∠LMN = ∠L\) [From (3)]
\(∠M = ∠L\) -----(6)
From (4) and (5), we get
\(∠LMN + ∠NME = ∠L + ∠NEM\)
\(∠LMN = ∠L\) [From (3)]
\(∠M = ∠L\) -----(6)
Proof for (ii): \(∠N = ∠O\)
\(LM || NO\) and \(LO\) is a transversal.
\(∠L + ∠O = 180^°\) ------(7) []
Similarly, \(∠M + ∠N = 180^°\) -------(8)
From (7) and (8), we get
\(∠L + ∠O = ∠M + ∠N\)
\(∠N = ∠O\) [From (6)]
From (7) and (8), we get
\(∠L + ∠O = ∠M + ∠N\)
\(∠N = ∠O\) [From (6)]
Proof for (iii): \(∆LMN ≅ ∆MLO\)
In \(∆LMN\) and \(∆MLO\), we have
\(LM = ML\) [Common]
\(MN = LO\) [Given]
\(∠LMN = ∠MLO\) [Proved]
\(LM = ML\) [Common]
\(MN = LO\) [Given]
\(∠LMN = ∠MLO\) [Proved]
Hence, \(∆LMN ≅ ∆MLO\) []
Proof for (iv): Diagonal \(LN =\) Diagonal \(MO\)
Since, \(∆LMN ≅ ∆MLO\) [Proved]
\(LN = MO\) []