MCQ
In a quadrilateral $A B C D$, which is not a trapezium, it is known that $\angle D A B=\angle A B C=60^{\circ}$. Moreover, $\angle C A B=\angle C B D$. Then,
- A$A B=B C+C D$
- B$A B=A D+C D$
- ✓$A B=B C+A D$
- D$A B=A C+A D$
✓
Answer
Correct option: C.
$A B=B C+A D$
c
(c)
(c)
$A B C D$ is a quadrilateral
$\angle D A B=\angle A B C=60^{\circ}$
and $\quad \angle C A B=\angle C B D$
Construction, $A D$ and $B C$ produced to meet at such that
$\triangle A E B$ is an equilateral.
$\because \quad A B=B E=A E$
In $\triangle B D E$ and $\triangle A B C$,
$\angle B E D=\angle A B C$
$\angle D B E=\angle C A B$ given,
$[\because \angle D B E=\angle D B C]$
$\triangle B E D \sim \triangle A B C$
$\frac{B E}{A B}=\frac{B D}{A C}=\frac{E D}{B C} \Rightarrow \frac{B E}{A B}=\frac{E D}{B C}$
$\frac{A B}{A B}=\frac{A E-A D}{B C}$
$\Rightarrow \quad A E-A D=B C \Rightarrow A B=A D+B C$
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