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.

               

Answers (1)

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  
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 \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}

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