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6. $\Delta ABC$ is an isosceles triangle in which $AB=AC$ . Side BA is produced to D such that $AD=AB$ (see Fig.). Show that $\angle BCD$ is a right angle.

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Consider $\Delta$ ABC,
It is given that AB = AC

So, $\angle ACB = \angle ABC$ (Since angles opposite to the equal sides are equal.)

Similarly in $\Delta$ ACD,

We have AD = AB
and $\angle ADC = \angle ACD$
So,

$\angle CAB + \angle ACB + \angle ABC = 180^{\circ}$

$\angle CAB\ +\ 2\angle ACB = 180^{\circ}$
or $\angle CAB\ = 180^{\circ}\ -\ 2\angle ACB$ ...........................(i)

And in $\Delta$ ADC,
$\angle CAD\ = 180^{\circ}\ -\ 2\angle ACD$ ..............................(ii)

Adding (i) and (ii), we get :
$\angle CAB\ +\ \angle CAD\ = 360^{\circ}\ -\ 2\angle ACD\ -\ 2\angle ACB$

or $180^{\circ}\ = 360^{\circ}\ -\ 2\angle ACD\ -\ 2\angle ACB$

and $\angle BCD\ =\ 90^{\circ}$

Posted by

Devendra Khairwa

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