AB is a diameter of a circle. `AH and BK` are perpendiculars from A and B respectively to the tangent at P Prove that `AH + BK = AB`.

Explanation

Let `AH=x, BK=y and BM=z`
`triangleMBK~triangleMAH" "`(AA corollary)
`:." "(BK)/(AH)=(BM)/(AM)" "implies" "(y)/(x)=(z)/(2r+z)`
`:." "2ry+yz=xz" "implies" "z(x-y)=2ry`
`implies" "z=(2ry)/(x-y)" "...(1)`
Similarly,
`DeltaMBK~DeltaMOP" "`(AA corollary)
`:." "(BK)/(OP)=(BM)/(OM)" "implies" "(y)/(r)=(z)/(z+r)`
`implies" "yz+yr=zr`
`implies" "z(r-y)=yr`
`implies" "z=(yr)/(r-y)" "...(2)`
From (1) and (2), we get
`(2ry)/(x-y)=(yr)/(r-y)" "implies" "(2)/(x-y)=(1)/(r-y)`
`implies" "2r-2y=x-y`
`implies" "x+y=2r`
`AH+BK=AB` Hence Proved.