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

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 \DeltaABC, 
It is given that   AB = AC

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

Similarly in \DeltaACD, 

We have  AD = AB 
and  \angle ADC = \angle ACD                                          ..............................(i)

And in \DeltaADC  \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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