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Show that

i) \(\angle{A}\) = \(\angle{B}\)

ii) \(\angle{C}\) = \(\angle{D}\)

iii) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\)

iv) Diagonal AC = Diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].

Answer :

Given: ABCD is trapezium.

AB || CD and AD = BC

Now, extend AB and draw a line through C parallel to DA intersecting AB produced at E.

Now, ADCE is a parallelogram.

\(\therefore \) AD || CE and AD = CE

But AD = BC

\(\therefore \) AD = BC = CE

i) We know that, \(\angle{A}\) + \(\angle{E}\) = \(180^\circ\)

(Since, interior angles on the same side of the transversal )

\(\Rightarrow \) \(\angle{E}\) = \(180^\circ\) - \(\angle{A}\)

\(\therefore \) \(\angle{E}\) = \(180^\circ\) - \(\angle{A}\)

(Since, BC = EC)

Also, \(\angle{ABC}\) = \(180^\circ\) - \(\angle{CBE}\)

(Since, ABE is straight line)

\(\angle{ABC}\) = \(180^\circ\) - \(180^\circ\) + \(\angle{A}\)

\(\angle{B}\) = \(\angle{A}\)

Hence, it is proved.

ii)Now, \(\angle{A}\) + \(\angle{D}\) = \(180^\circ\)

(Since, interior angles on the same side of the transversal)

\(\Rightarrow \) \(\angle{D}\) = \(180^\circ\) - \(\angle{A}\)

\(\Rightarrow \) \(\angle{D}\) = \(180^\circ\) - \(\angle{B}\)

(from eq.(i)) ...(ii)

Also, \(\angle{C}\) + \(\angle{B}\) = \(180^\circ\) ...180 (Since, interior angles on
the same side of the transversal BC)

\(\Rightarrow \) \(\angle{C}\) = \(180^\circ\) - \(\angle{B}\) ...(iii)

from Eq. (ii) and (iii), we get,

\(\angle{C}\) = \(\angle{D}\)

Hence, proved.

iii)Now, in \(\triangle{ABC}\) and \(\triangle{BAD}\), we have,

AD = BC ...(Given)

\(\angle{A}\) = \(\angle{B}\) ...(From Eq.(i))

AB = BA ...(Common side)

\(\therefore \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\) ...(By SAS rule)

Hence, proved.

iv)

\(\because \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\)

\(\therefore \) AC = BD

Hence, it is proved.

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