Question

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

Answer 1

Hint: As per given data, we can draw a diagram that is
  

Now, AB is the diameter of the circle. O is the centre of the circle and AB passes through O. AH and BK are perpendiculars drawn from A and B on tangent P. P is the tangent of the circle. AH and BK are the perpendiculars drawn from the extremities of the diameter AB. As per the diagram ABKH is a rectangle because AH and BK are perpendiculars. So, from the diagram AB is equal to HK (opposite sides of the rectangle) and we also know that lengths of the tangents drawn from an external point to the circle are equal. And then adding the above equalities the proof for the above question will be found.

Complete step-by-step answer:
As per given data,
Perpendiculars drawn from A and B on tangent at P are AH and BK.
As AH and BK are perpendiculars
Therefore, ABKH is a rectangle
So, AB=HK (opposite sides of rectangle are equal) $...\left( 1 \right)$
We know that the length of the tangents drawn from an external point to a circle are equal.
So,
AH=HP $...\left( 2 \right)$
BK=PK $...\left( 3 \right)$
Now, adding Equation $\left( 2 \right)$ and $\left( 3 \right)$,
AH+BK=HP+PK=HK
From equation $\left( 1 \right)$
AH+BK=AB.
Hence, it is proved that when AH and BK are perpendiculars from A and B respectively to the tangent at P then AH+BK=AB.

Note: Perpendiculars always make ${90^ \circ }$ to the base and with that it is possible to say that ABKH is a rectangle. It is to be remembered that tangents from an external point to a circle are of equal length.