Question

Construct an angle of ${90}^o$ at the initial point of a given ray and justify the construction.

Answer 1

Steps of construction

1. Draw a line segment OA.
2. Taking O as center and any radius, draw an arc cutting OA at B.
3. Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.
4. With C as center and the same radius, draw an arc cutting the arc at D.
5. With C and D as center and radius more than $\frac{1}{2}$CD, draw two arc intersecting at P.
6. Join OP.
7. Thus, $\angle AOP={90}^o$

$\underset{–}{Justification}$

Join OC and BC

Thus,

$OB=BC=OC$ [Radius of equal arcs]

$\therefore △OCB$ is an equilateral triangle

$\therefore \angle BOC={60}^o$

Join OD,OC and CD

Thus,

$OD=OC=DC$ [Radius of equal arcs]

$\therefore △DOC$ is an equilateral triangle

$\therefore \angle DOC={60}^o$

Join PD and PC

Now,

In $△ODP$ and $△OCP$

$OD=OC$ [Radius of same arcs]

$DP=CP$ [Arc of same radii]

$OP=OP$ [Common]

$\therefore △ODP\cong △OCP$ [SSS congruency]

$\therefore \angle DOP=\angle COP$ [CPCT]

So, we can say that

$\angle DOP=\angle COP=\frac{1}{2}\angle DOC$

$\angle DOP=\angle COP=\frac{1}{2}\times 60={30}^o$

Now,

$\angle AOP=\angle BOC+\angle COP$

$\angle AOP=60+30$

$\angle AOP={90}^o$

Hence justified.