The triangle ABCis inscribed in a circle, AX, BY and CZ of the angles `angle BAC, angle ABC and angle ACB`, intersect at the point X,Y,Z on the circle respectively. Prove that AX is perpendicular to XZ.

Explanation

Let `Delta ABC` is inscribed in the circle will centre at O. AX, BY and CZ are the bisectors of `angle BAC, angle ABC and angle ACB` respectively and intersect the circle at X,Y and Z respectively.
Let AX,BY and CZ intersect each other at O. Then O is the incentre of the circle. Again let AX intersect YZ at the point E.
To prove : We hve to prove that `AX bot YZ i., AE bot YZ`
`:.` are AY are CY[ by the converet of theorem-35]
`:.` similarly, are BX= are CX and arc AX= BZ.
[ `:.` angles in the same segment of a cricle are equal]
Again, are AZ and are BZ produced two equal central angles.
`:. angleAOX= angle BOZ.......(1)`
Similary `angle AOY= angle COY[ :.` AY = are CY]
`or, angle AOY= angleBOZ`[ oppoiste angles]....(2)
`:. angle AOZ= angleAOY or, angleEOZ= angleEOY .......(3)`
Bow, in `Delta EOY and Delta EOZ, OY= OZ ( :.` radii of same circle,)
`angle EOY= angle EOZ` [ from (3) ] and OE is common to both.

`:. Delta EPO~= Delta EOZ` [ by the S-A-S conditino of congreuency]
`angle OEY= angle OEZ[ :.` similar angles of congruent triangles]
But these two angles are adjustment obtained when the line segment OE stands upon the line segment YZ and they are equal.
`:.` each is right angle i.e., ` angle OEY= angle OEZ`= right angle.
`:. OE bot or AX bot YZ` ltbr `:.` Ax is perpendicular to YZ