Oman
^@ D^@, ^@E^@ and ^@F^@ are respectively the midpoints of sides ^@AB^@, ^@BC^@ and ^@CA^@ of ^@\Delta ABC^@. Find the ratio of the areas of ^@\Delta DEF^@ and ^@\Delta ABC^@.
Answer:
^@1:4^@
- In ^@\Delta ABC^@, ^@D^@ and ^@F^@ are the midpoints of sides ^@AB^@ and ^@CA^@ respectively.
Therefore, ^@DF || BC^@ [ By midpoint theorem ]
^@\implies\space\space DF || BE.^@
Similarly, ^@\space\space EF || BD^@.
Therefore, ^@ BEFD ^@ is a parallelogram.
^@\implies \angle B = \angle EFD, EF = BD = \dfrac { 1 } { 2 } AB ^@ and ^@ DF = BE = \dfrac { 1 } { 2 } BC^@.
Also, ^@ ECFD ^@ is a parallelogram.
^@\implies\space\space \angle EDF = \angle C.^@ - Now, in ^@\Delta DEF^@ and ^@\Delta CAB^@, we have @^ \begin{aligned} &\angle EFD = \angle B \\ and \space\space& \angle EDF = \angle C \\ \therefore \space\space& \Delta DEF \sim \Delta CAB &&[\text{ By AA-similarity}] \\ \text{ So, }& \dfrac { ar(\Delta DEF ) } { ar( \Delta ABC ) } = \dfrac {ar( \Delta DEF )} { ar(\Delta CAB) } = \dfrac { DF^2 } { BC^2 } = \dfrac{ (\dfrac{1} { 2 } BC)^2 } { BC^2 } = \dfrac { 1 } { 4 } \end{aligned} @^ Thus, the ratio of the areas of ^@\Delta DEF^@ and ^@\Delta ABC^@ is ^@1:4^@.
