The two circles with centre X and Y intersect each other at the points A and B, A joined with the mid -point S of XY and the perpendicular on SA through the point A is drawn which intersect the two circles at the point P and Q .Prove that PA = AQ.

Explanation

Let us draw two perpendicular XM and YN on PA and AQ respectively. Let us join X, A and Y, A
`because XM bot PA , therefore AM = (1)/(2) PA ………. (1)`
Again `YM bot PA, therefore AN = (1)/(2) AQ ………. (2)`
Now , `because ` S is the mid-point XY.
`therefore AS bot XY [ because ` the line segment joining two centres of two circles bisect their common chord in right angles ] .
`therefore ` in right - angled AXS and AYX.
we get , XS = YS [ ` because` S is the mid - point of XY.
`therefore angle ASX = angle ASY [ because ` each is right angle .
and AS in common to both .
`therefore Delta AXS ~ = Delta AYS , therefore XA = YA [ because` similar sides of two congrument triangles ]
Now , XY XY, and PQ are both perpendiculars to PQ.

`therefore XY || PQ`
Again ` because ` XM and YN are both perpendicula r to Pq .
` therefore XM = YM`
Now in `Delta AXM and Delta AYN,XY = YA, SX = SY` and SA is common to both
` therefore Delta AXM ~= Delta AYM ` [ by the S-S -S condition of congruent of congruency]
`therefore` AM = AN [` because`similar sides of congruent triangles ]
` or (1)/(2) PA = (1)/(2) AQ` [ from (1) and (2)]
` or PA = AQ , therefore PA = AQ `