Step 1: Convert the given information into a figure


                    ABC right triangle

                ABD, BCE and AFC are equilateral triangles

            To Prove:  area (∆ABD) + area (∆BCE) = area (∆ACF)

Step 2: Search for the similar triangles in the assumed figure.

            NOTE: Triangle ABD & Triangle ACF are equilateral triangle in

            an  equilateral triangle each angle us equal to 60 degree

              In ABD angle A= angle B = angle D= 60 degree

              In ACF angle A= angle C = angle F = 60 degree

                    Therefore ABD ~ ACF

                    Similarly  BCE ~ ACF

Step 3: Find the areas of the similar triangles

            THEOREM:  The ratio of the areas of two similar triangles is equal to the

                          ratio of the squares of their corresponding sides

                         \frac{Area\ \triangle ABC}{Area\ \triangle ACF} =(\frac{AB}{AC})^2 ...................(1)

                         \frac{Area\ \triangle BCE}{Area\ \triangle ACF} =(\frac{BC}{AC})^2 .................................(2)

Step 3: Solve the equations mentioned above

              Add equation (1) and (2)

            \frac{Area \triangle ABC}{Area\ \triangle ACF} + \frac{Area \triangle BCE}{Area \triangle ACF} = (\frac{AB}{AC})^2 + (\frac{BC}{AC})^2


Step 4: Simplify R.H.S and L.H.S sides to prove the required equation.


(\frac{AB}{AC})^2 + (\frac{BC}{AC})^2 = (\frac{(AB)^2}{(AC})^2) + (\frac{(BC)^2}{(AC)^2})

                          = (\frac{(AB)^2+(BC)^2}{\left((AC\right)^2})

                          = (\frac{(AC)^2}{\left((AC\right)^2}) (since Pythagoras)

                          = 1


\frac{Area \triangle ABC}{Area\ \triangle ACF} + \frac{Area \triangle BCE}{Area \triangle ACF} = \frac{Area\ \ \triangle BCE+Area\ \ \triangle ABC}{Area\ \ \triangle ACF}

L.H.S = R.H.S

\frac{Area\ \triangle BCE+Area\ \triangle ABC}{Area\ \triangle ACF} = 1

Area triangle BCE + area triangle ABC = Area \triangle ACF

Hence proved