Question

Construct an angle of $90^\circ $ at the initial point of a given ray and justify the construction.

Answer 1

Hint:We will draw two arcs of $60^\circ $ and $120^\circ $ then the bisector will be of angle $90^\circ $.

Complete step-by-step solution
Let us start the problem by step-by-step construction.The following are the steps to construct an angle of $90^\circ $.
Step 1:
Draw a ray OA. Taking O as centre and any radius, draw an arc that cuts OA at B.

Step 2:
Now, taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.

Step 3: With C as centre and the same radius, draw an arc cutting the arc at D.
Step 5:
With C and D as centres and radius more than half of CD, draw two arcs intersecting at P.

Step 6:
Join OP.

Therefore, the angle of $90^\circ $ at the initial point of the given ray is obtained.

JUSTIFICATION
Here, we will prove that$\angle AOP = 90^\circ $.

Construction
Join OC and BC
Also, sides OB, BC and OC are equal because these are radii of equal arcs.
$OB = BC = OC$
Therefore, \[\Delta OCB\] is an equilateral triangle in which,
\[\angle BOC = 60^\circ \]
Now, Join OD, OC and CD.
Also, sides OD, OC and DC are equal because these are radii of equal arcs.
\[OD = OC = DC\]
Therefore, \[\Delta DOC\] is an equilateral triangle in which,
\[\angle DOC = 60^\circ \]
Also, Join PD and PC.

In triangle ODP and triangle OCP,
\[\begin{array}{*{20}{l}}
{OC = OD}\\
{DP = CP}\\
{OP = OP}
\end{array}\]
Therefore, both the triangles $\Delta ODP \cong \Delta OCP$ are congruent through the SSS congruent rule.
Also, by CPCT,
$\angle DOP = \angle COP$
Now, we can say that
$\begin{array}{c}
\angle DOP = \angle COP\\
\angle DOP = \dfrac{1}{2} \times \angle DOC\\
\angle DOP = \dfrac{1}{2} \times 60^\circ = 30^\circ
\end{array}$
Since, we know that,
$\begin{array}{l}
\angle AOP = \angle BOC + \angle COP\\
\angle AOP = 60^\circ + 30^\circ \\
\angle AOP = 90^\circ
\end{array}$
Hence, $\angle AOP = 90^\circ $ is justified.

Note:Circumcircle of a triangle is a circle on which all three vertices of triangle lie. Make sure to draw the circle through the points A, B and P and join line OB and OC.