If in the given figure OR ' = 2 × OR , OS ' = 2 × OS and OT ' = 2 × OT . Prove that △ R S T ~ △ R ' S ' T ' .

Given: In ∆ O S ' R ' and ∆ O S R , ∠ S ' O R ' = ∠ S O R (Common angle) O S ' O S = O R ' O R = 2 1 . . . . ( i ) Therefore, by SAS criterion ∆ O S ' R ' ~ ∆ O S R Similarly, in ∆ O S ' T ' and ∆ O S T ∠ S ' O T ' = ∠ S O T (Common angle) O S ' O S = O T ' O T = 2 1 . . . . ( ii ) Therefore, by SAS criterion ∆ O S ' T ' ~ ∆ O S T From ( i ) and ( ii ) , we get O S ' O S = O R ' O R = O T ' O T ⇒ ∆ R ' S ' T ' ~ ∆ R S T

In triangle ACB , ∠ ACB = 90 ° and CD ⊥ AB . Prove that BC 2 AC 2 = BD AD .

Given: In ∆ C A B , ∠ A C B = 90 ° and C D ⊥ A B We know that if a perpendicular drawn from the vertex of the right angle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other. ⇒ ∆ A C D ~ ∆ A B C . . . . . ( i ) and ∆ B C D ~ ∆ B A C . . . . . ( ii ) From ( i ) , ∆ A C D ~ ∆ A B C If two triangles are similar, then the ratio of their corresponding sides are equal. ⇒ A C A B = A D A C ⇒ A C 2 = A B × A D . . . . . ( iii ) Similarly, from ( ii ) , ∆ B C D ~ ∆ B A C ⇒ B C B A = B D B C ⇒ B C 2 = B A × B D . . . . ( iv ) Dividing ( iv ) by ( iii ) , we get B C 2 A C 2 = B A × B D A B × A D ⇒ B C 2 A C 2 = B D A D Hence, proved