Question

In the adjoining figure AD bisects ∠A. Arrange AB, BD, and DC in the descending order of their heights.

Answer 1

Given: $\Delta ABC,AD$ bisects $\angle A$,
In $\Delta ABC$,
Sum of angles $={180}^o$
$\angle A+\angle B+\angle C={180}^o$
$\angle A+{60}^o+{40}^o={180}^o$
$\angle A={80}^o$
$\angle BAD=\angle DAC={40}^o$

$\angle A={80}^o,\angle C={40}^o$
Since, $\angle A>\angle C$
$BC>AB$ (Sides opposite greater angles is greater ) .. (i)

In $\Delta ADC$
$\angle ACD=\angle DAC={40}^o$
Thus, $AD=DC$ (Isosceles triangle property)

Now, In $\Delta ABD$
Sum of angles $={180}^o$
$\angle ABD+\angle ADB+\angle BAD={180}^o$
$60+\angle ADB+{40}^o={180}^o$
$\angle ADB={80}^o$

$\angle ABD={60}^o$ and $\Delta ADB={80}^o$
Since, $\angle ABD>\angle ADB$
Thus, $AD>AB$
or $DC>AB$ (Since ,$AD=DC)$ ... (ii)
and we know $BC=BD+DC$
Hence, $BC>DC$ ... (iii)
Hence, from (i), (ii) and (iii)
$BC>DC>AB$