MHB Can AC^2 be Proven to Equal AB(AB+BC) in Triangle ABC with Angles B=80 and C=40?

[Albert1]

$\triangle ABC,\angle B=80^o,\angle C=40^o$
$prove :\overline{AC}^2=\overline{AB}(\overline{AB}+\overline{BC})$

 

[kaliprasad]

$\triangle ABC,\angle B=80^o,\angle C=40^o$
$prove :\overline{AC}^2=\overline{AB}(\overline{AB}+\overline{BC})$

Spoiler

we have $\angle A=60^\circ$

using law of sines we have

$\frac{\overline{BC}+\overline{AB}}{\overline{AC}}= \frac{\sin \angle A + \sin \angle C}{\sin \angle B}=\frac{\sin \,60^\circ + \sin \,40^\circ}{\sin \,80^\circ} $
$= 2\frac{\sin \,50^\circ \cos \,10^\circ}{\cos 10^\circ} = 2\sin \,50^\circ $
further
$\frac{\overline{AC}}{\overline{AB}}= \dfrac{\sin\,80^\circ}{\sin\,40^\circ}= \dfrac{2\sin\,40^\circ\cos\,40^\circ}{\sin\,40^\circ}= 2\cos\,40^\circ = 2 \sin \, 50^\circ$

from above 2 we get the result

 

[Albert1]

$\triangle ABC,\angle B=80^o,\angle C=40^o$
$prove :\overline{AC}^2=\overline{AB}(\overline{AB}+\overline{BC})$

Spoiler

hope someone can prove it using geometry

 

[kaliprasad]

Spoiler

hope someone can prove it using geometry

Spoiler

Draw the triangle ABC and extend AB ro D such that BC = BD
join CD

now triangle CBD is isosceles triangle and hence $\overline{BC} = \overline{BD}$
now $\triangle ABC$ and $\triangle ACD$ are similar

so $\frac{\overline {AB}}{\overline {AC}}= \frac{\overline {AC}}{\overline {AD}} $
or ${\overline {AB}} * {\overline {AD}}= ({\overline {AC}})^2 $
or ${\overline {AB}} * ({\overline {AB} + \overline {BD}})= ({\overline {AC}})^2 $
or ${\overline {AB}} * ({\overline {AB} + \overline {BC}})= ({\overline {AC}})^2 $