Question

Que: Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.​

Answer 1

Answer:

Solution;

To Prove,

To prove: A circle drawn with Q as centre, will pass through A, B and O (i.e. QA = QB = QO)

Since all sides of a rhombus are equal,

AB = DC

Now, multiply (½) on both sides

(½)AB = (½)DC

So, AQ = DP

⇒ BQ = DP

Since Q is the midpoint of AB,

AQ= BQ

Similarly,

RA = SB

Again, as PQ is drawn parallel to AD,

RA = QO

Now, as AQ = BQ and RA = QO we have,

QA = QB = QO (hence proved).

Hope it will be helpful :)☺️

Answer 2

❤ hOlA mAtE ❤

To Prove ,

==> To prove : A circle drawn with Q as centre , will pass through A , B and O ( i.e. QA = QB = QO )

==> Since all sides of a rhombus are equal , AB = DC

==> Now , multiply ( 12 ) on both sides ( 12 ) AB = ( 1/2 ) DC

==> So , AQ = DP → BQ = DP

==> Since Q is the midpoint of AB ,

==> AQ = BQ

==> Similarly , RA = SB

==> Again , as PQ is drawn parallel to AD , RA = QO

==> Now , as AQ = BQ and RA = QO we have , QA = QB = QO ( hence proved ) .

Hope it helps you ✔️