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# ABCD is a trapezium in which AB || CD and AD = BC (see figure). 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].

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.

$$\therefore$$ AD || CE and AD = CE
$$\therefore$$ 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,
$$\angle{A}$$ = $$\angle{B}$$ ...(From Eq.(i))
$$\therefore$$ $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{BAD}$$ ...(By SAS rule)
$$\because$$ $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{BAD}$$
$$\therefore$$ AC = BD