Question

In the given figure, quadrilateral ABCD is a square.
triangle BCE on side BC and triangle ACF on the diagonal AC are similar to each other. Show that A(triangle BCE) = 1/2 A (triangle ACF)​

Answer 1

Answer

Since ABCD is a square,

AB=BC=CD=DA  and AC=  √2 BC [Diagonal=  √2 ]

​△BCE∼△ACF

Area(△ACF)  / Area(△BCE)     =  BC^2 / AC^2

                                                = BC^2 / (√2BC)^2

                                                = 1 /2

Area(△BCE)=   1/2 Area(△ACF)   [Taking the denominator to RHS]

Hence, showed.