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$ \mathrm{ABCD} $ is a trapezium in which $ \mathrm{AB} \| \mathrm{CD} $ and $ \mathrm{AD}=\mathrm{BC} $ (see below figure). Show that
(i) $ \angle \mathrm{A}=\angle \mathrm{B} $
(ii) $ \angle \mathrm{C}=\angle \mathrm{D} $
(iii) $ \triangle \mathrm{ABC} \equiv \triangle \mathrm{BAD} $
(iv) diagonal $ \mathrm{AC}= $ diagonal $ \mathrm{BD} $
[Hint: Extend $ \mathrm{AB} $ and draw a line through $ \mathrm{C} $ parallel to $ \mathrm{DA} $ intersecting $ \mathrm{AB} $ produced at E.]
"
Given:
\( \mathrm{ABCD} \) is a trapezium in which \( \mathrm{AB} \| \mathrm{CD} \) and \( \mathrm{AD}=\mathrm{BC} \).
To do :
We have to show that
(i) \( \angle \mathrm{A}=\angle \mathrm{B} \)
(ii) \( \angle \mathrm{C}=\angle \mathrm{D} \)
(iii) \( \triangle \mathrm{ABC} \equiv \triangle \mathrm{BAD} \)
(iv) diagonal \( \mathrm{AC}= \) diagonal \( \mathrm{BD} \)
Solution :
Extend \( \mathrm{AB} \) and draw a line through \( \mathrm{C} \) parallel to \( \mathrm{DA} \) intersecting \( \mathrm{AB} \) produced at $E$.
(i) $ADCE$ is a parallelogram.
$CE = AD$ (Opposite sides of a parallelogram are equal)
$AD = BC$ (Given)
This implies,
$BC = CE$
$\Rightarrow \angle CBE = \angle CEB$
$\angle EAD+\angle CEB = 180^o$
$\angle EAD+\angle CBE = 180^o$ ($\angle CBE = \angle CEB$)
$\angle CBA +\angle CBE = 180^o$ (Linear pair)
This implies,
$\angle A = \angle B$
(ii) $\angle A+\angle D =180^o$ (Adjacent angles of a parallelogram are supplementary)
$\angle B+\angle C = 180^o$ (Angles on the same side of transversal are supplementary)
$\angle A+\angle D = \angle B+\angle C$
$\angle A+\angle D = \angle A+\angle C$ ($\angle A = \angle B$)
This implies,
$\angle D = \angle C$
(iii) In $\triangle ABC$ and $\triangle BAD$,
$AB = AB$ (Common side)
$\angle DBA = \angle CBA$
$AD = BC$ (Given)
Therefore, by SAS congruency, we get,
$\triangle ABC \cong \triangle BAD$
(iv) $\triangle ABC \cong \triangle BAD$
This implies,
$AC = BD$ (CPCT)
Therefore,
diagonal \( \mathrm{AC}= \) diagonal \( \mathrm{BD} \).
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