Proof : In ∆ABD and ∆CDE,
AD = DE Â Â Â [by construction]
∠ADB = ∠CDE
[vertically opposite angles]
and    BD = DC    [AD is a median]
Therefore, by using SAS congruent condition
        Â
          [by CPCT]
Similarly, we can prove
         Â
                [by CPCT]
It is given that:
  Â
     Â
Therefore, by using SSS congruent condition
         Â
                  ...(i)
Similarly, Â Â Â Â Â Â Â Â Â Â Â Â Â ...(ii)
Adding (i) and (ii), we get
∠1 + ∠3 = ∠2 + ∠4
∠A = ∠P
Now, in ∆ABC and ∆PQR
          Â
and      Â
Therefore, by using SAS similar condition
∆ABC ~ ∆PQR Hence Proved.
It is given that:
∆ABC ~ ∆DEF
We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
∴   Â
       Â
       Â
         Â
          Â
Hence, BC = 11.2 cm.
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.