Question

In a quadrilateral ABCD,which is not a trapezium,it is know that $\angle DAB=\angle ABC=60$. Moreover,$\angle CAB=\angle CBD$. Then:

• A. AB=BC+CD
• B. AB=AD+CD
• C. AB=BC+AD
• D. AB=AC+AD

Answer 1

The correct option is C AB=BC+AD

In quadrilateral $ABCD$,

$\angle DAB=\angle ABC={60}^o$ [ Given ]

$\angle CAB=\angle CBD$ [ Given ]

$\Rightarrow$ $BC$ and $AD$ are produced to intersect at $E.$

In $△ABE,$

$\angle EBA=\angle EAB={60}^o$

$\therefore$ $\angle BEA={60}^o$ [ Since, sum of angles of triangle is ${180}^o$ ]

$\therefore$ $△ABE$ is an equilateral triangle.

$\Rightarrow$ $AB=BE=EA$ ---- ( 1 )

In $△ABC$ and $△BED$

$\Rightarrow$ $\angle BED=\angle CBA={60}^o$

$\Rightarrow$ $\angle EBD=\angle CAB$ [ Given ]

$\Rightarrow$ $BE=AB$ [ From ( 1 ) ]

$\therefore$ $△ABC\cong △BED$ [ By $ASA$ congruence rule ]

$\Rightarrow$ $DE=BC$ [ Corresponding sides of congruent triangles ] --- ( 2 )

$\Rightarrow$ $AB=AE$

$\Rightarrow$ $AB=AD+DE$

$\Rightarrow$ $AB=AD+BC$ [ From ( 2 ) ]