Question

In the given figure, $AD$ bisects $\angle A$. Arrange $AB$, $BC$ and $DC$ in ascending order.

• A. $BC>DC>AB$
• B. $BC<DC<AB$
• C. $BC>AB>DC$
• D. $BC<AB<DC$

Answer 1

The correct option is A $BC>DC>AB$

Given; $△ABC$, $AD$ bisects $\angle A$,
In $△ABC$,
$\angle A={70}^{\circ }$, $\angle C={35}^{\circ }$
Since, $\angle A>\angle C$
$BC>AB$ (Sides opposite greater angles is greater) (1)
In $△ADC$
$\angle ACD=\angle DAC={35}^{\circ }$
Thus, $AD=DC$ (Isosceles triangle property)
Now, In $△ABD$
$\angle ABD={75}^{\circ }$ and $△ADB={70}^{\circ }$
Since, $\angle ABD>\angle ADB$
Thus, $AD>AB$
or $DC>AB$ (Since, $AD=DC$) (2)
and we know $BC=BD+DC$
Hence, $BC>DC$ (3)
Hence, from (1), (2) and (3)
$BC>DC>AB$