Question

prove that the circle drawn with any side of a rhombus as a diameter, passes through the point of intersection of its diagonals

Answer 1

it can be done by contra positive proof assume anothee point with which it passes now angle in a semicircle is 90

diagonal of rhombus intersect at 90degree hence the pt. with which circle pass should coincide with pt. of intersection

Answer 2

Correct question:-

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

$\huge\sf\star{\pink{\underline{\underline{\mathrm{Explanation:}}}}}$

______________________________

To prove:

$\leadsto$ 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

$\hookrightarrow$ (½)AB = (½)DC

$\leadsto$ So, AQ = DP

$\hookrightarrow$ BQ = DP

$\leadsto$ Since Q is the midpoint of AB,

$\hookrightarrow$ AQ= BQ

Similarly,

$\hookrightarrow$ RA = SB

$\leadsto$ Again, as PQ is drawn parallel to AD,

RA = QO

$\leadsto$ Now, as AQ = BQ and RA = QO we have,

$\huge\star$$\boxed{\purple{QA=QB=QO}}$

$\huge\star{\green{\underline{\mathrm{Hence,Proved!}}}}$