Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Given:△ABC∼△DEF
O is a median of BC and P is a median of EF
To Prove:A(△ABC)/A(△DEF)=(AO)2/(DP)2
Proof:Since,△ABC∼△DEF
∴∠A=∠D,∠B=∠E,∠C=∠F(Corresponding Angles of Similar Triangles)(1)
Also,AB/DE=BC/EF=AC/DF(Corresponding Sides of Similar Triangles). (2)
Since,BC=2BO and EF=2EP
∴Equation (2) can be written as,AB/DE=BC/EF=AC/DF=BO/EP. (3)
In△AOB and △DPE
∠B=∠E(From 1)
AB/DE=BO/EP(From 3)
∴By SAS Criterion of Similarity,△AOB∼△DPE
∴AB/DE=BO/EP=AO/DP=Ratio of their heights(4)(Corresponding Sides of Similar Triangles)
A(△ABC)/A(△DEF)=(1/2×BC×Height)/(1/2×EF×Height)=(AO)2/(DP)2