Two chords AB and CD of a circle are perpendicular to each other. If perpendicular drawn to AD from the point of intesection of those two chords AB and CD is produced to meet BC at the point E prove that the point E is the mid- point of BC.

Explanation

`AB bot CD, :. angle = angle BPC=90^(@)`
`:. angle ADB= 90^(@)- angle DAP .....(1)`
Now, `angle ADC= angle ABC............(2)`
Again, `angle BPE= angle APF[ :.` opposite angles] .........(3)
Now, `angle APF=90^(@)-anglePAC[ :. angle AFP=90^(@)]` ltbgt `= angle ADP[ :. angleAPD= 90^(@)........(4)`
from (3) and (4) we get `angle BPE= angle ADP= angleADC= angle ABC= angle PBE`
`:. "in" Delta BPE= angle PBE, :. BE=PE .......(5)`
Similary it can be proved that `CE=PE..........(6)`
Then from (5) and (6) we get BE=CE
Hence E is the mid-point of BC.