Question

Sides AB and AC of a $△$ ABC are equal. BC is produced through C upto point D such that AC = CD.

D and A are joined and produced (through vertex A) upto point E.

If $\angle$ BAE = ${108}^0$; find $\angle$ ADB.

Answer 1

In $\Delta$ ABD,

$\angle BAE=\angle 3+\angle ADB$

108= $\angle 3+\angle ADB$

But AB = AC

$\angle 3=angle2$

108= $\angle 2+\angle ADB$……(i)

Now,

In $\Delta$ ACD,

$\angle 2=\angle 1+\angle ADB$

But AC = CD

$\angle 1=\angle ADB$

$\angle 2=\angle ADB+\angle ADB$

$\angle 2=2\angle ADB$

Putting this value in (i)

108 = 2$\angle ADB+\angle ADB$

3$\angle ADB={108}^0$

$\angle ADB={36}^0$