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Solution :

If we draw a line `EF` parallel to `AB` in the given figure,<br> it will divide rectangle `ABCD` into two rectangles `ABFE` and `CDEF`.<br>
Then, `AE = BF` and `CF = DE->(1)`<br>
Now, in `Delta OEA`,<br>
`OA^2 = OE^2+AE^2->(2)`<br>
In `Delta OCF`,<br>
`OC^2 = OF^2+CF^2->(3)`<br>
In `Delta OBF`,<br>
`OB^2 = OF^2+BF^2->(4)`<br>
In `Delta OED`,<br>
`OD^2 = OE^2+DE^2->(5)`<br>
Now, adding (2) and (3),<br>
`OA^2+OC^2 = OE^2+AE^2+OF^2+CF^2`<br>
From(1), `AE = BF` and `CF = DE`<br>
So,<br>
`OA^2+OC^2 = (OE^2+DE^2)+(OF^2+BF^2)`<br>
`OA^2+OC^2 = OB^2+OD^2`<br>
**What we already learned in previous classes**

**How similarity is different from congruence.**

**Similar Polygons**

**Similar Triangles and their properties**

**Basic proportionality Theorem or Thales Theorem - If a line is drawn parallel to one side of a triangle intersecting the other two sides; then it divides the two sides in the same ratio.**

**If in a `DeltaABC`; a line DE||BC; intersects AB in D and AC in E; Then `AB`/
`AD` = `AC`/`AE`**

**Converse of Basis proportionality theorem : If a line divides any two sides of a triangle in the same ratio; then the line must be parallel to the third side.**

**The internal angle bisector of an angle of a triangle divide the opposite side internally in the ratio of the sides containgthe angle**

**If a line through one vertex of a triangle divides the opposite sides in the Ratio of other two sides; then the line bisects the angle at the vertex.**

**The external angle bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.**