Question
✓
Answer
Construction: Draw a line through C parallel to DA intersecting AB produced at E.
- CE = AD (Opposite sides of a parallelogram)
AD = BC (Given)
Therefor, BC = CE
also,
$\angle\text{A}+\angle\text{CBE}=180^{0}$ (Angles on the same side of transversal and ∠CBE = ∠CEB)
$\angle\text{B}+\angle\text{CBE}=180^{0}$ (Linear pair)
$\angle\text{A}=\angle\text{B}$
- $\angle\text{A}+\angle\text{D}=\angle\text{B}+\angle\text{C}=180^{0}$ (Angles on the same side of transversal)
$\angle\text{A}+\angle\text{D}=\angle\text{B}+\angle\text{C}$
$\angle\text{D}=\angle\text{C}$
- In $\angle\text{ABC}$ and $\angle\text{BAD}$
AB = AB (Common)
$\angle\text{DBA}=\angle\text{CBA}$
AD = BC (Given)
- Thus, by SAS $\angle\text{DBA}=\angle\text{CBA}$ Congruence condition.
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