Question

If two chords AB and CD of a circle perpendicular to each other at point O then prove that `OA^2+OB^2+OC^2+OD^2=4R^2?`

Answer 1

Please see below.

Explanation:

Let us consider the chords `AB` and `CD` intersecting each other at right angle as shown below. Join `AC,BC,BD` and `AD`. Let us construct diameter `AP` and join `PD`.

Now as `DeltaAOD` is right angled, `OA^2+OD^2=AD^2`

and as `DeltaBOC` is right angled, `OB^2+OC^2=BC^2`

i.e. `OA^2+OB^2+OC^2+OD^2=AD^2+BC^2`

Now consider right angled `DeltasADP` and `AOC`.

As `/_APD` and `/_ACO` are subtended by chord `AD` on the circle, `/_APD=/_ACO`. Also `/_AOC=/_ADP=90^@`.

Hence `DeltasADP` and `AOC` are similar

and hence `(AD)/(AP)=(AO)/(AC)` and as `AP=2R`, `AC=2Rxx(AO)/(AD)`

Similarly we can draw diameter `BQ`, join `QC` and have similar `DeltasBQC` and `BOD` and then

`(BD)/(BO)=(BQ)/(BC)` and as `BQ=2R`, `BD=2Rxx(BO)/(BC)`

Hence, `OA^2+OB^2+OC^2+OD^2=AD^2+BD^2`

= `(2Rxx(AO)/(AD))^2+(2Rxx(BO)/(BC))^2`

= `4R^2(((AO)/(AD))^2+((BO)/(BC))^2)`

Now in `DeltasAOD` and `BOC`, as they are similar, we have

`(AO)/(AD)=(OC)/(BC)`

and hence #`OA^2+OB^2+OC^2+OD^`

= `4R^2(((OC)/(BC))^2+((BO)/(BC))^2)`

= `4R^2((OC^2+OB^2)/(BC^2))`

= `4R^2`

Answer 2

Let `AB and CD` be two mutually perpendicular chords of a circle of center K.

If `R` represents the length of its radius , we are to prove that

`color (red)(OA^2+OB^2+OC^2+OD^2=4R^2)`

Constructions

`AX "||"CD` is drawn, which intersects the circle at X . A perpendicular `XY` is drawn on `CD` at `Y`

`Band C;Cand A;Aand D;BandX;Dand X` are joined

As `AO_|_CD and AX"||"CD, /_XAB=90^@` Hence `BX` will be diameter of the circle i.e `BX=2R`

Now `DeltasAOD and BOC` are similar

So `(OC)/(OA)=(OB)/(OD)`

`=>OA*OB=OC*OD`

`ACDX` trapezium is cyclic. So `/_ACO=/_XDY`

In `DeltaAOCand DeltaXYD`

`/_ACO=/_XDY`

`/_AOC=/_XYD=90^@`

`AO=XY` (opposite sides of the rectangle `AXYO)`

Hence `DeltaAOC~= DeltaXYD->color(red)(OC=YD)`

Now applying Pythagoras theorem for `DeltaBAX` we get

`AB^2+AX^2=BX^2`

`=>(OA+OB)^2+OY^2=(2R)^2`

`=>(OA+OB)^2+(OD-YD)^2=(2R)^2`

`=>(OA+OB)^2+(OD-OC)^2=4R^2`

`=>OA^2+OB^2+2OA*OB+OC^2+OD^2-2OC*OD=4R^2`

`=>OA^2+OB^2+cancel((2OC*OD))+OC^2+OD^2-cancel((2OC*OD))=4R^2`

`color (red)(=>OA^2+OB^2+OC^2+OD^2=4R^2)`
Proved