prove that a circle drawn by taking a side of a rhombus as diameter passes through point of intersection diago

Answer 1

ok, did it. now what do you want me to do.

Answer 2

Given:
Rhombus ABCD
AC and BD intersect at X
Circle E with diameter AB.

Prove:
X is on circle E


-->statements:

1) Assume X is not on circle E

2) m angle AXB ≠ 90

3) AC is perpendicular to BD

4) m angle AXB = 90

5) X is on circle E


-->Reasons
1) assumption

2) If the vertex of an angle whose rays pass through the endpoints of a diameter of a circle is contained in that circle, then the angle is a right angle.

3) the diagonals of a rhombus are perpendicular

4) definition of perpendicular

5) assumption is false because of contradiction of statements 3 and 4

^_^

Answer 3

Proof:
assume a rhombus PQRS
let pq be the diameter of the circle.
now diagonals pr and qs intersect at A.
according to the definition of rhombus,diagonals of rhombus intersect at right angles.
so PAQ=90 . . . . . . . . . . .. . . ..1
NOW IF WE OBSERVE THE FIGURE THEN YOU'LL FIND THAT A LIES ON THE CIRCLE
becoz the diameter of circle always subtends a right angle,if the point A has to lie on the circle then anglePAQ=90,WHICH WE HAVE PROVED(FRM I)............2
FRM STATEMENTS 1 AND 2 A LIES ON THE CIRCLE.

Answer 4

The diagonals of a rhombus intersect at right angles. . . . .. .
or the the point of intersection of diagonals of a rhombus subtends a right angle at the side which is now the diameter
and hence it is on the circle

Answer 5

the answer is Q.

Answer 6

ok i did it
amnt i great?