G.159 circumscribe a square about a circle — Heath

IV.7

About a given circle to circumscribe a square.

Let ( A B C D ) be the given circle; thus it is required to circumscribe a square about the circle ( A B C D ).

Let two diameters / A C /, / B D / of the circle ( A B C D ) be drawn at right angles to one another, and through the points A, B, C, D let / F G /, / G H /, / H K /, / K F / be drawn touching the circle ( A B C D ). [G.118]

Then, since / F G / touches the circle ( A B C D ), and / E A / has been joined from the centre E to the point of contact at A,

[G.120]therefore the angles at A are right.

For the same reason

the angles at the points B, C, D are also right.

Now, since the angle < A E B ) is right, and the angle < E B G ) is also right,

[G.54]therefore / G H / is parallel to / A C /.

For the same reason

/ A C / is also parallel to / F K /,

so that / G H / is also parallel to / F K /. [G.57]

Similarly we can prove that

each of the straight lines / G F /, / H K / is parallel to / B E D /.

Therefore < F G H K >, < G H C A >, < A C K F >, < F G B D >, < B H K D > are parallelograms; therefore / G F / is equal to / H K /, and / G H / to / F K /. [G.62]

And, since / A C / is equal to / B D /, and / A C / is also equal to each of the straight lines / G H /, / F K /,

[G.62]while / B D / is equal to each of the straight lines / G F /, / H K /,

therefore the quadrilateral < F G H K > is equilateral.

I say next that it is also right-angled.

For, since < G B E A > is a parallelogram, and the angle < A E B ) is right, therefore the angle < A G B ) is also right. [G.62]

Similarly we can prove that

the angles at < G H K ), < H K F ), < K F G ) are also right.

Therefore < F G H K > is right-angled.

But it was also proved equilateral;

therefore it is a square;

and it has been circumscribed about the circle ( A B C D ).

Therefore about the given circle a square has been circumscribed.

Q. E. F.

Dependency Graph

Required for

G.563, G.565, G.570, G.571, G.572, G.573