Quadrilaterals — Maths STD 9 — Question

1. $\angle$A = $\angle$B
2. $\angle$C = $\angle$D
3. $\triangle$ABC $\cong$  $\triangle$BAD
4. diagonal AC = diagonal BD.

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

Proof :

1. AB || CD . . . [Given]
   AD || ED . . .[By construction]
   $\therefore$ AECD is a parallelogram . . . [A quadrilateral is a parallelogram if a pair of Opp. sides are parallel and of equal length]
   $\therefore$ AD = EC . . .[Opp. sides of a || gm are equal]
   But AD = BC . . .[Given]
   $\therefore$ EC = BC
   $\therefore$ $\angle$CBE = $\angle$CEB . . . [$\angle$s opposite to equal side of a triangle are equal] . . .(1)
   $\angle$B + $\angle$CBE = 180^o . . .[Linear pair axiom]. . . (2)
   As AD || EC . . .[By construction]
   and transversal AE intersect them
   $\angle$A + $\angle$CEB = 180^o . . . [The sum of interior angles on the same side of the transversal is 180^o] . . . (3)
   From (2) and (3),
   $\angle$B + $\angle$CBE = $\angle$A + $\angle$CEB
   But $\angle$CBE = $\angle$CEB . . .[From (1)]
   $\therefore$ $\angle$B = $\angle$A or $\angle$A = $\angle$B
2. As AB || CD
   $\therefore$ $\angle$A + $\angle$D = 180^o. . .[The sum of interior angles on the same side of the transversal is 180^o]
   and $\angle$B + $\angle$C = 180^o
   $\therefore$ $\angle$A + $\angle$D = $\angle$B + $\angle$C
   But $\angle$A = $\angle$B . . .[As proved in (i)]
   $\therefore$ $\angle$D = $\angle$C or $\angle$C = $\angle$D
3. In $\triangle$ABC and  $\triangle$BAD,
   AB = BA . . . [Common]
   BC = AD . . .[Given]
   $\angle$ABC = $\angle$BAD . . .[From (i)]
   $\therefore$ $\triangle$ABC $\cong$ $\triangle$BAD . . .[SAS rule]
4. As $\triangle$ABC $\cong$ $\triangle$BAD . . [From (iii)]
   $\therefore$ AC = BD . . .[c.p.c.t.]