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Quadrilaterals — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsQuadrilaterals4 Marks
Question
ABCD is a trapezium in which AB || CD and AD = BC. Show that,
  1. $\angle\text{A}=\angle\text{B}$
  2. $\angle\text{C}=\angle\text{D}$
  3. $\triangle\text{ABC}\cong\triangle\text{BAD}$
  4. Diagonal AC = diagonal BD.
✓

Answer

Construction: Draw a line through C parallel to DA intersecting AB produced at E.

  1. 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)

$\angle\text{A}=\angle\text{B}$

  1. $\angle\text{A}+\angle\text{D}=\angle\text{B}+\angle\text{C}=180^{0}$ (Angles on the same side of transversal)

$\angle\text{A}+\angle\text{D}=\angle\text{B}+\angle\text{C}$

$\angle\text{D}=\angle\text{C}$

  1. In $\angle\text{ABC}$ and $\angle\text{BAD}$

AB = AB (Common)

$\angle\text{DBA}=\angle\text{CBA}$

AD = BC (Given)

  1. Thus, by SAS $\angle\text{DBA}=\angle\text{CBA}$ Congruence condition.

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