O is the orthocentre of `Delta ABC and AD bot BC`. If AD is produced it interect the circumcircle of `Delta ABC` at the point G. Prove that `OD= DG`.

Explanation

The perpendiculars AD , BE aned CF on the sides BC,CA and AB respectively of `Delta ABC` intersect each other at So, O is the orthocentre of `Delta ABC`. When Adis extended it intersects as the circumcircle with centre P at the poiint G.
To prove : We have provet that OD= DG.
Construction : Let us join C,G
Proof : `angle BAD and angle BCG` are two anlges in circle, produced by the are `overset(frown)(BG)` .
`:. angle BAG=angle BCG` [ by theorem-35]
or, `90^(@)- angle ABD= angle BCG [ :. AD bot BC, :. angle ADB=90^(@)]`
`or,90^(@)- (90^(@)-angle BCF)= angle BCG [ :. CF bot AB. :. angle BFC= 90^(@)]`
`or, angle BCF= angle BCG`
`or, angle OCD= angle GCD[ :. angle BCF and OCD and angle OCD ` also `angle BCG and angle GCD` are same angles]
`:. angle OCD= angle GCD......(1)`
Now, in `DeltaCOD and Delta CGD`,
`angle CDO= angle GCD[ :.` each is right angle]
` angle OCD= angle GCD` [ from (1)]
and CD is common to both.
`:. Delta COD~= Delta CGD` [ by the A-A-S codition of congreuency]
`:. OD=DG[ :.` simila side os fongruent triangles]
Hence OD= DG