A B C is an isosceles triangle in which A B = A C . If B A ¯ is produced to D so that A D = A B , prove that ∠ B C D = 90 ° .

Given, A B C is an isosceles triangle in which A B = A C . If B A ¯ is produced to D so that A D = A B . In ∆ A B C , as A B = A C , ∠ A B C = ∠ A C B And in ∆ A C D , as A C = A D , ∠ A C D = ∠ A D C And as A B = A C = A D , ∠ A B C = ∠ A C B = ∠ A C D = ∠ A D C . We know that, sum of angles of a triangle is 180 ° . So, in ∆ B C D , ∠ B C D + ∠ C B D + ∠ B D C = 180 ° ⇒ ∠ B C D + ∠ B C A + ∠ A C D = 180 ° ⇒ ∠ B C D + ∠ B C D = 180 ° ⇒ 2 ∠ B C D = 180 ° ⇒ ∠ B C D = 90 ° Hence, proved.

A B C is an isosceles triangle in which A B = A C . If B A ¯ is produced to D so that A D = A B , prove that ∠ B C D = 90 ° .

Given, A B C is an isosceles triangle in which A B = A C . If B A ¯ is produced to D so that A D = A B . In ∆ A B C , as A B = A C , ∠ A B C = ∠ A C B And in ∆ A C D , as A C = A D , ∠ A C D = ∠ A D C And as A B = A C = A D , ∠ A B C = ∠ A C B = ∠ A C D = ∠ A D C . We know that, sum of angles of a triangle is 180 ° . So, in ∆ B C D , ∠ B C D + ∠ C B D + ∠ B D C = 180 ° ⇒ ∠ B C D + ∠ B C A + ∠ A C D = 180 ° ⇒ ∠ B C D + ∠ B C D = 180 ° ⇒ 2 ∠ B C D = 180 ° ⇒ ∠ B C D = 90 ° Hence, proved.