Question

The line joining the foot of perpendicular drawn from a point lying on the circumcircle. Of a triangle to the sides of a triangle is a straight line.

Answer 1

Given : `DeltaABC` and P is any on the cirucmcircle of `Delta ABC`.PL,PM and PN are the perpendicular on BC, AC and AB respectively.
To Prove : LMN is a straight line.
Construction : Join PA and PC.
Proof: square `ANPM` is cyclic .
`(because angle PNA+anglePMA+90^(@)+90^(@)=180^(@))`
`angle1=angle4`
(exterior angle of a cyclic quadrilateral is equal to opposite interior angle).....(2)
`rArr angle1 =angle5` [from (1) and (2)]....(3)
Now, since `angle2=angle6` (each `90^(@)`)
and these are the angles subtended by line joining two points lying on the same side of line.
So, square `MPCL`is cyclic.
`therefore (angle2+angle3)+angle5=180^(@)` (sum of opposite angles of a cyclic quadrilateral)
`rArrangle 2+angle3+angle 1=180^(@)` [from (3)]
`rArrangle1 +angle2+angle 3=180^(@)`
i.e., LMN is straight line.
This straight line LMN is known as the "Simpson line".