Question

If the diagonal BD of a quadrilateral ABCD bisects both $\angle B$ and $\angle D$ then,
AB$=$AD

• A. True
• B. False

Answer 1

The correct option is B False

Given: $AB=AC$
$\angle ABC=\angle ACB$ (Isoceles triangle property).. (II)
Hence, $180-\angle ABC=180-\angle ACB$
$\frac{1}{2}(180-\angle ABC)=\frac{1}{2}(180-\angle ACB)$
$\angle DBC=\angle DCB$ ...(I)
Adding I and II
$\angle ABC+\angle DBC=\angle ACB+\angle DCB$
$\angle DBA=\angle DCA$... (III)
Now, In $△ABD$ and $△ACD$
$AB=AC$ (Given)
$\angle DBA=\angle DCA$ (From III)
$AD=AD$ (Common side)
Thus,$△ABD\cong △ACD$ (SAS rule)
Hence, $\angle DAB=\angle DAC$ (AD bisects $\angle BAC$)