Question

In the given figure, $AD$ bisects $\angle A$. If $\angle B={60}^0$, $\angle C={40}^0$, then arrange $AB$, $BD$ and $DC$ in ascending order of their lengths.

• 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 $△ABC$,
Sum of angles = 180
$\angle A+\angle B+\angle C=180$
$\angle A+60+40=180$
$\angle A={80}^{\circ }$
$\angle BAD=\angle DAC={40}^{\circ }$
$\angle A={80}^{\circ }$, $\angle C={40}^{\circ }$
Since, $\angle A>\angle C$
$BC>AB$ (Sides opposite greater angles is greater) (1)
In $△ADC$
$\angle ACD=\angle DAC={40}^{\circ }$
Thus, $AD=DC$ (Isosceles triangle property)
Now, In $△ABD$
Sum of angles = 180
$\angle ABD+\angle ADB+\angle BAD=180$
$60+\angle ADB+40=180$
$\angle ADB={80}^{\circ }$
$\angle ABD={60}^{\circ }$ and $△ADB={80}^{\circ }$
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)
$BC>DC>AB$