Prove-that-inside-a-square-a-semi-circle-which-touches-all-the-four-sides-of-the-square-is-possible-with-ruler-and-compass-

Question Number 4080 by Rasheed Soomro last updated on 27/Dec/15

Prove that inside a square, a semi - circle,

which touches all the four sides of the

square, is possible with ruler and compass .

Commented by Rasheed Soomro last updated on 29/Dec/15

A Semi - circle in a square touching

all the sides . There is a method by

ruler and compass, which I have

forgotten at the moment . However

you can get idea from the above

image .

Answered by Rasheed Soomro last updated on 30/Dec/15

First I write ruler and compass method of

the required figure, then I^{'} ll prove the method .

Pl see the image

ABCD is a given square . ∠ DBC has been

bisected . Let the bsector meets the side DC

at E .

∠ DBC is one of eight diagonal - side angles .

You can select any angle to bisect it .

From E a line ∥ BC has been drawn that

met diagonal BD at F . F is center of required

semi - circle . Through F a line segment GH

parallel to diagonal AC has been drawn . GH is

diameter of the semi - circle . Taking center F

and radius equal to FG or FH or FE, required

semi - circle has been drawn .

PROOF ▸ Without the loss of generality, {let}^{'} s

assume that side of the given square is unity

∴ Each diagonal is \sqrt{2} .

Semi - circle touches all the sides of square .

{We}^{'} ll prove its being touched for two consecutive

sides BC and CD in this proof Touching other

two sides is easy to prove .

△ BEC is aright angled triangle, in which

∠ BCE = π / 2, ∠ EBC = π / 8 and BC = 1

\tan ∠ EBC = EC / BC \Rightarrow EC = \tan (π / 8) = \sqrt{2} - 1

So, EC = \sqrt{2} - 1

Now DE = CD - EC \Rightarrow DE = 1 - (\sqrt{2} - 1) = 2 - \sqrt{2}

Or DE = 2 - \sqrt{2}

△ DEF is a right angled triangle, in which

∠ FDE = π / 4, ∠ DFE = π / 4 (∠ DFE is complement

of ∠ FDE)

∵ Two angles are equal

∴ EF = DE = 2 - \sqrt{2} \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots . I

\sin ∠ FDE = EF / DF \Rightarrow \sin (π / 4) = (2 - \sqrt{2}) / DF

\frac{1}{\sqrt{2}} = \frac{2 - \sqrt{2}}{DF} \Rightarrow DF = 2 \sqrt{2} - 2

Now BF = BD - DF = \sqrt{2} - (2 \sqrt{2} - 2) = 2 - \sqrt{2}

BF = 2 - \sqrt{2}

In △ BFG, ∠ FBG = ∠ FGB

∴ GF = BF = 2 - \sqrt{2} \dots \dots \dots \dots \dots \dots \dots \dots II

From I & II

EF = GF = 2 - \sqrt{2}

Semi - circle touches DC and BC

QED

Answered by Rasheed Soomro last updated on 29/Dec/15

Commented by Rasheed Soomro last updated on 30/Dec/15

Actually the paper was white and new but

the photo seems as old as if it were taken

two centuries ago! So your advice is very

right .

Commented by Yozzii last updated on 29/Dec/15

I f {y o u}^{'} d l i k e t o m a k e p i c t u r e s l i k e

t h e s e m o r e d e f i n e d w i t h a m o r e w h i t e

b a c k g r o u n d, t r y u s i n g t h e a p p c a l l e d

C a m S c a n n e r . {T h a t}^{'} s w h a t I u s e d t o

t o p r o d u c e t h e p i c t u r e s f o r s o m e

q u e s t i o n s I s h a r e d r e c e n t l y .

Commented by RasheedSindhi last updated on 30/Dec/15

Thanks . I^{'} ll try .