Textbook - Constructions Questions in English

(N/A) Given a triangle $ABC$,we are required to construct another triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle $ABC$.
Steps of Construction:
$1.$ Draw any ray $BX$ making an acute angle with $BC$ on the side opposite to the vertex $A$.
$2.$ Locate $4$ (the greater of $3$ and $4$ in $\frac{3}{4}$) points $B_1, B_2, B_3$ and $B_4$ on $BX$ so that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$3.$ Join $B_4C$ and draw a line through $B_3$ (the $3rd$ point,$3$ being smaller of $3$ and $4$ in $\frac{3}{4}$) parallel to $B_4C$ to intersect $BC$ at $C'$.
$4.$ Draw a line through $C'$ parallel to the line $CA$ to intersect $BA$ at $A'$.
Then,$\Delta A'BC'$ is the required triangle.
Let us now see how this construction gives the required triangle.
Since $B_3C' \parallel B_4C$,by Basic Proportionality Theorem,$\frac{BC'}{C'C} = \frac{BB_3}{B_3B_4} = \frac{3}{1}$.
Therefore,$\frac{BC}{BC'} = \frac{BC' + C'C}{BC'} = 1 + \frac{C'C}{BC'} = 1 + \frac{1}{3} = \frac{4}{3}$,i.e.,$\frac{BC'}{BC} = \frac{3}{4}$.
Also,$C'A' \parallel CA$. Therefore,$\Delta A'BC' \sim \Delta ABC$.
So,$\frac{A'B}{AB} = \frac{A'C'}{AC} = \frac{BC'}{BC} = \frac{3}{4}$.

(N/A) Given a triangle $ABC$,we are required to construct a triangle whose sides are $\frac{5}{3}$ of the corresponding sides of $\Delta ABC$.
Steps of Construction:
$1.$ Draw any ray $BX$ making an acute angle with $BC$ on the side opposite to the vertex $A$.
$2.$ Locate $5$ points (the greater of $5$ and $3$ in $\frac{5}{3}$) $B_{1}, B_{2}, B_{3}, B_{4}$ and $B_{5}$ on $BX$ so that $BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}$.
$3.$ Join $B_{3}$ (the $3$rd point,$3$ being smaller of $3$ and $5$ in $\frac{5}{3}$) to $C$ and draw a line through $B_{5}$ parallel to $B_{3}C$,intersecting the extended line segment $BC$ at $C^{\prime}$.
$4.$ Draw a line through $C^{\prime}$ parallel to $CA$ intersecting the extended line segment $BA$ at $A^{\prime}$.
Then $\Delta A^{\prime}BC^{\prime}$ is the required triangle.
For justification of the construction,note that $\Delta ABC \sim \Delta A^{\prime}BC^{\prime}$.
Therefore,$\frac{AB}{A^{\prime}B} = \frac{AC}{A^{\prime}C^{\prime}} = \frac{BC}{BC^{\prime}}$.
But,$\frac{BC}{BC^{\prime}} = \frac{BB_{3}}{BB_{5}} = \frac{3}{5}$.
So,$\frac{BC^{\prime}}{BC} = \frac{5}{3}$,and therefore,$\frac{A^{\prime}B}{AB} = \frac{A^{\prime}C^{\prime}}{AC} = \frac{BC^{\prime}}{BC} = \frac{5}{3}$.

(N/A) line segment of length $7.6 \, cm$ can be divided in the ratio of $5: 8$ as follows:
$1.$ Draw a line segment $AB$ of $7.6 \, cm$ and draw a ray $AX$ making an acute angle with line segment $AB$.
$2.$ Locate $13 (= 5 + 8)$ points,$A_1, A_2, A_3, \dots, A_{13}$,on $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = \dots = A_{12}A_{13}$.
$3.$ Join $BA_{13}$.
$4.$ Through the point $A_5$,draw a line parallel to $BA_{13}$ (by making an angle equal to $\angle AA_{13}B$) at $A_5$ intersecting $AB$ at point $C$.
$C$ is the point dividing line segment $AB$ of $7.6 \, cm$ in the required ratio of $5: 8$.
The lengths of $AC$ and $CB$ can be measured. They are $2.9 \, cm$ and $4.7 \, cm$ respectively.
Justification:
By construction,we have $A_5C \parallel A_{13}B$. By applying the Basic Proportionality Theorem $(BPT)$ for the triangle $AA_{13}B$,we obtain:
$\frac{AC}{CB} = \frac{AA_5}{A_5A_{13}}$
From the construction,$AA_5$ contains $5$ equal divisions and $A_5A_{13}$ contains $8$ equal divisions.
Therefore,$\frac{AA_5}{A_5A_{13}} = \frac{5}{8}$.
Thus,$\frac{AC}{CB} = \frac{5}{8}$. This justifies the construction.

(N/A) $1.$ Draw a line segment $AB = 4 \, cm$. Taking point $A$ as centre,draw an arc of $5 \, cm$ radius. Similarly,taking point $B$ as its centre,draw an arc of $6 \, cm$ radius. These arcs will intersect each other at point $C$. Now,$AC = 5 \, cm$ and $BC = 6 \, cm$,and $\triangle ABC$ is the required triangle.
$2.$ Draw a ray $AX$ making an acute angle with line $AB$ on the opposite side of vertex $C$.
$3.$ Locate $3$ points $A_1, A_2, A_3$ on line $AX$ such that $AA_1 = A_1A_2 = A_2A_3$.
$4.$ Join $BA_3$ and draw a line through $A_2$ parallel to $BA_3$ to intersect $AB$ at point $B'$.
$5.$ Draw a line through $B'$ parallel to the line $BC$ to intersect $AC$ at $C'$.
$\triangle AB'C'$ is the required triangle.
The construction can be justified by proving that $\frac{AB'}{AB} = \frac{B'C'}{BC} = \frac{AC'}{AC} = \frac{2}{3}$.
By construction,we have $B'C' \parallel BC$.
$\therefore \angle AB'C' = \angle ABC$ (Corresponding angles).
In $\triangle AB'C'$ and $\triangle ABC$,
$\angle AB'C' = \angle ABC$ (Proved above),
$\angle B'AC' = \angle BAC$ (Common angle).
$\triangle AB'C' \sim \triangle ABC$ ($AA$ similarity criterion).
$\Rightarrow \frac{AB'}{AB} = \frac{B'C'}{BC} = \frac{AC'}{AC}$ .......... $(1)$
In $\triangle AA_2B'$ and $\triangle AA_3B$,
$\angle A_2AB' = \angle A_3AB$ (Common angle),
$\angle AA_2B' = \angle AA_3B$ (Corresponding angles).
$\therefore \triangle AA_2B' \sim \triangle AA_3B$ ($AA$ similarity criterion).
$\Rightarrow \frac{AB'}{AB} = \frac{AA_2}{AA_3} = \frac{2}{3}$ ......... $(2)$
From equations $(1)$ and $(2)$,we obtain $\frac{AB'}{AB} = \frac{B'C'}{BC} = \frac{AC'}{AC} = \frac{2}{3}$.
This justifies the construction.

(N/A) $1.$ Draw a line segment $AB$ of $5\, cm$. Taking $A$ and $B$ as centers,draw arcs of $6\, cm$ and $7\, cm$ radius respectively. Let these arcs intersect each other at point $C$. $\triangle ABC$ is the required triangle having side lengths of $5\, cm$,$6\, cm$,and $7\, cm$ respectively.
$2.$ Draw a ray $AX$ making an acute angle with line $AB$ on the opposite side of vertex $C$.
$3.$ Locate $7$ points,$A_1, A_2, A_3, A_4, A_5, A_6, A_7$ (as $7$ is greater between $5$ and $7$) on line $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7$.
$4.$ Join $BA_5$ and draw a line through $A_7$ parallel to $BA_5$ to intersect the extended line segment $AB$ at point $B'$.
$5.$ Draw a line through $B'$ parallel to $BC$ intersecting the extended line segment $AC$ at $C'$. $\triangle AB'C'$ is the required triangle.
Justification:
The construction can be justified by proving that $AB' = \frac{7}{5} AB$,$B'C' = \frac{7}{5} BC$,and $AC' = \frac{7}{5} AC$.
In $\triangle ABC$ and $\triangle AB'C'$:
$\angle ABC = \angle AB'C'$ (Corresponding angles)
$\angle BAC = \angle B'AC'$ (Common)
$\triangle ABC \sim \triangle AB'C'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = \frac{5}{7}$ (Equation $1$)
In $\triangle AA_5B$ and $\triangle AA_7B'$:
$\angle A_5AB = \angle A_7AB'$ (Common)
$\angle AA_5B = \angle AA_7B'$ (Corresponding angles)
$\triangle AA_5B \sim \triangle AA_7B'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{AB'} = \frac{AA_5}{AA_7} = \frac{5}{7}$ (Equation $2$)
On comparing equations $1$ and $2$,we obtain:
$\frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = \frac{5}{7}$
$\Rightarrow AB' = \frac{7}{5} AB, B'C' = \frac{7}{5} BC, AC' = \frac{7}{5} AC$.
This justifies the construction.

(N/A) Let us assume that $\triangle ABC$ is an isosceles triangle having $CA = CB$,base $AB = 8\, cm$,and altitude $AD = 4\, cm$ (where $D$ is the midpoint of $AB$).
$A$ $\triangle AB'C'$ whose sides are $\frac{3}{2}$ times those of $\triangle ABC$ can be drawn as follows:
$1.$ Draw a line segment $AB = 8\, cm$. Construct the perpendicular bisector of $AB$ to find the midpoint $D$. Let the perpendicular line be $OO'$.
$2.$ Taking $D$ as the centre,draw an arc of radius $4\, cm$ on the perpendicular line to intersect at point $C$. Join $AC$ and $BC$ to form the isosceles $\triangle ABC$.
$3.$ Draw a ray $AX$ making an acute angle with $AB$ on the side opposite to vertex $C$.
$4.$ Locate $3$ points $A_1, A_2,$ and $A_3$ on $AX$ such that $AA_1 = A_1A_2 = A_2A_3$.
$5.$ Join $BA_2$. Draw a line through $A_3$ parallel to $BA_2$ to intersect the extended line segment $AB$ at $B'$.
$6.$ Draw a line through $B'$ parallel to $BC$ to intersect the extended line segment $AC$ at $C'$. $\triangle AB'C'$ is the required triangle.
Justification:
Since $A_2B \parallel A_3B'$,by Basic Proportionality Theorem in $\triangle AA_3B'$,we have $\frac{AB}{AB'} = \frac{AA_2}{AA_3} = \frac{2}{3}$. Thus,$AB' = \frac{3}{2}AB$.
Since $BC \parallel B'C'$,$\triangle ABC \sim \triangle AB'C'$. Therefore,$\frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = \frac{2}{3}$,which implies $B'C' = \frac{3}{2}BC$ and $AC' = \frac{3}{2}AC$. This justifies the construction.

(A) $\angle B = 45^{\circ}, \angle A = 105^{\circ}$
The sum of all interior angles in a triangle is $180^{\circ}$.
$\angle A + \angle B + \angle C = 180^{\circ}$
$105^{\circ} + 45^{\circ} + \angle C = 180^{\circ}$
$\angle C = 180^{\circ} - 150^{\circ} = 30^{\circ}$
The required triangle can be drawn as follows:
$1.$ Draw a $\triangle ABC$ with side $BC = 7 \, cm$,$\angle B = 45^{\circ}$,$\angle C = 30^{\circ}$.
$2.$ Draw a ray $BX$ making an acute angle with $BC$ on the opposite side of vertex $A$.
$3.$ Locate $4$ points $B_1, B_2, B_3, B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$4.$ Join $B_3C$. Draw a line through $B_4$ parallel to $B_3C$ intersecting the extended $BC$ at $C'$.
$5.$ Through $C'$,draw a line parallel to $AC$ intersecting the extended line segment $BA$ at $A'$. $\triangle A'BC'$ is the required triangle.
Justification:
In $\triangle ABC$ and $\triangle A'BC'$,
$\angle ABC = \angle A'BC'$ (Common)
$\angle ACB = \angle A'C'B$ (Corresponding angles)
$\triangle ABC \sim \triangle A'BC'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{A'B} = \frac{BC}{BC'} = \frac{AC}{A'C'} \dots(1)$
In $\triangle BB_3C$ and $\triangle BB_4C'$,
$\angle B_3BC = \angle B_4BC'$ (Common)
$\angle BB_3C = \angle BB_4C'$ (Corresponding angles)
$\triangle BB_3C \sim \triangle BB_4C'$ ($AA$ similarity criterion)
$\Rightarrow \frac{BC}{BC'} = \frac{BB_3}{BB_4} = \frac{3}{4} \dots(2)$
From $(1)$ and $(2)$,we get $\frac{AB}{A'B} = \frac{BC}{BC'} = \frac{AC}{A'C'} = \frac{3}{4}$.
Thus,$A'B = \frac{4}{3} AB$,$BC' = \frac{4}{3} BC$,and $A'C' = \frac{4}{3} AC$. This justifies the construction.

(N/A) It is given that sides other than hypotenuse are of lengths $4 \,cm$ and $3 \,cm$. Clearly,these will be perpendicular to each other.
The required triangle can be drawn as follows:
$1.$ Draw a line segment $AB = 4 \,cm$. Draw a ray $AX$ making $90^{\circ}$ with it at $A$.
$2.$ Draw an arc of $3 \,cm$ radius while taking $A$ as its centre to intersect the ray at $C$. Join $BC$. $\triangle ABC$ is the required triangle.
$3.$ Draw a ray $AY$ making an acute angle with $AB$,opposite to vertex $C$.
$4.$ Locate $5$ points $A_1, A_2, A_3, A_4, A_5$ on $AY$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5$.
$5.$ Join $A_3B$. Draw a line through $A_5$ parallel to $A_3B$ intersecting the extended line segment $AB$ at $B'$.
$6.$ Through $B'$,draw a line parallel to $BC$ intersecting the extended line segment $AC$ at $C'$. $\triangle AB'C'$ is the required triangle.
Justification:
The construction can be justified by proving that $AB' = \frac{5}{3} AB, B'C' = \frac{5}{3} BC, AC' = \frac{5}{3} AC$.
In $\triangle ABC$ and $\triangle AB'C'$,
$\angle ABC = \angle AB'C'$ (Corresponding angles)
$\angle BAC = \angle B'AC'$ (Common)
$\therefore \triangle ABC \sim \triangle AB'C'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'}$ .......$(1)$
In $\triangle AA_3B$ and $\triangle AA_5B'$,
$\angle A_3AB = \angle A_5AB'$ (Common)
$\angle AA_3B = \angle AA_5B'$ (Corresponding angles)
$\therefore \triangle AA_3B \sim \triangle AA_5B'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{AB'} = \frac{AA_3}{AA_5} = \frac{3}{5}$ .........$(2)$
On comparing equations $(1)$ and $(2)$,we obtain
$\frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = \frac{3}{5}$
$AB' = \frac{5}{3} AB, B'C' = \frac{5}{3} BC, AC' = \frac{5}{3} AC$
This justifies the construction.

(N/A) pair of tangents to the given circle can be constructed as follows:
$1.$ Taking any point $O$ of the given plane as centre,draw a circle of $6 \, cm$ radius. Locate a point $P$,$10 \, cm$ away from $O$.
Join $OP$.
$2.$ Bisect $OP$. Let $M$ be the mid-point of $PO$.
$3.$ Taking $M$ as centre and $MO$ as radius,draw a circle.
$4.$ Let this circle intersect the previous circle at points $Q$ and $R$.
$5.$ Join $PQ$ and $PR$. $PQ$ and $PR$ are the required tangents.
The lengths of tangents $PQ$ and $PR$ are $8 \, cm$ each.
Justification:
The construction can be justified by proving that $PQ$ and $PR$ are the tangents to the circle (whose centre is $O$ and radius is $6 \, cm$). For this,join $OQ$ and $OR$.
$\angle PQO$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.
$\therefore \angle PQO = 90^{\circ}$
$\Rightarrow OQ \perp PQ$
Since $OQ$ is the radius of the circle,$PQ$ has to be a tangent to the circle. Similarly,$PR$ is a tangent to the circle.

(N/A) Tangents on the given circle can be drawn as follows:
$1.$ Draw a circle of $4 \, cm$ radius with centre as $O$ on the given plane.
$2.$ Draw a circle of $6 \, cm$ radius taking $O$ as its centre. Locate a point $P$ on this circle and join $OP.$
$3.$ Bisect $OP.$ Let $M$ be the mid-point of $PO.$
$4.$ Taking $M$ as its centre and $MO$ as its radius,draw a circle. Let it intersect the given circle at the points $Q$ and $R$.
$5.$ Join $PQ$ and $PR.$ $PQ$ and $PR$ are the required tangents.
It can be observed that $PQ$ and $PR$ are of length $4.47 \, cm$ each.
In $\triangle PQO$:
Since $PQ$ is a tangent,$\angle PQO = 90^{\circ}$.
$PO = 6 \, cm$ (hypotenuse),
$QO = 4 \, cm$ (radius).
Applying the Pythagoras theorem in $\triangle PQO$,we obtain:
$PQ^2 + QO^2 = PO^2$
$PQ^2 + (4)^2 = (6)^2$
$PQ^2 + 16 = 36$
$PQ^2 = 36 - 16$
$PQ^2 = 20$
$PQ = \sqrt{20} = 2\sqrt{5} \approx 4.47 \, cm$.
Justification:
The construction can be justified by proving that $PQ$ and $PR$ are the tangents to the circle (whose centre is $O$ and radius is $4 \, cm$). For this,join $OQ$ and $OR$.
Since $\angle OQP$ is an angle in a semi-circle,$\angle OQP = 90^{\circ}$.
$\Rightarrow OQ \perp PQ$.
Since $OQ$ is the radius of the circle,$PQ$ must be a tangent to the circle. Similarly,$PR$ is a tangent to the circle.

(N/A) The tangents can be constructed on the given circle as follows:
$1.$ Taking any point $O$ on the given plane as the centre,draw a circle of $3 \, cm$ radius.
$2.$ Take one of its diameters and extend it on both sides. Locate two points $P$ and $Q$ on this diameter such that $OP = OQ = 7 \, cm$.
$3.$ Bisect $OP$ and $OQ$. Let $T$ and $U$ be the mid-points of $OP$ and $OQ$ respectively.
$4.$ Taking $T$ and $U$ as centres and $TO$ and $UO$ as radii,draw two circles. These two circles will intersect the original circle at points $V, W$ and $X, Y$ respectively. Join $PV, PW, QX,$ and $QY$. These are the required tangents.
Justification:
The construction can be justified by proving that $PV, PW, QX,$ and $QY$ are the tangents to the circle (whose centre is $O$ and radius is $3 \, cm$). For this,join $OV, OW, OX,$ and $OY$.
$\angle PVO$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.
$\therefore \angle PVO = 90^{\circ}$
$\Rightarrow OV \perp PV$
Since $OV$ is the radius of the circle,$PV$ must be a tangent to the circle.
Similarly,it can be shown that $PW, QX,$ and $QY$ are the tangents to the circle.

(N/A) The tangents can be constructed in the following manner:
$1.$ Draw a circle of radius $5\, cm$ with center $O$.
$2.$ Take a point $A$ on the circumference of the circle and join $OA$. Draw a perpendicular to $OA$ at point $A$.
$3.$ Draw a radius $OB$ making an angle of $120^{\circ} (180^{\circ} - 60^{\circ})$ with $OA$.
$4.$ Draw a perpendicular to $OB$ at point $B$. Let both the perpendiculars intersect at point $P$. $PA$ and $PB$ are the required tangents inclined at an angle of $60^{\circ}$.
Justification:
The construction can be justified by proving that $\angle APB = 60^{\circ}$.
By our construction:
$\angle OAP = 90^{\circ}$ (Tangent is perpendicular to the radius at the point of contact)
$\angle OBP = 90^{\circ}$ (Tangent is perpendicular to the radius at the point of contact)
And $\angle AOB = 120^{\circ}$
We know that the sum of all interior angles of a quadrilateral is $360^{\circ}$.
In quadrilateral $OAPB$:
$\angle OAP + \angle AOB + \angle OBP + \angle APB = 360^{\circ}$
$90^{\circ} + 120^{\circ} + 90^{\circ} + \angle APB = 360^{\circ}$
$300^{\circ} + \angle APB = 360^{\circ}$
$\angle APB = 360^{\circ} - 300^{\circ} = 60^{\circ}$
This justifies the construction.

(N/A) The tangents can be constructed on the given circles as follows:
$1.$ Draw a line segment $AB$ of $8 \, cm$. Taking $A$ and $B$ as centres,draw two circles of $4 \, cm$ and $3 \, cm$ radius respectively.
$2.$ Bisect the line $AB$. Let the mid-point of $AB$ be $C$. Taking $C$ as centre,draw a circle of radius $AC$ (or $BC$),which will intersect the circles at points $P, Q, R,$ and $S$. Join $BP, BQ, AS,$ and $AR$. These are the required tangents.
Justification:
The construction can be justified by proving that $AS$ and $AR$ are the tangents to the circle (whose centre is $B$ and radius is $3 \, cm$) and $BP$ and $BQ$ are the tangents to the circle (whose centre is $A$ and radius is $4 \, cm$). For this,join $AP, AQ, BS,$ and $BR$.
$\angle ASB$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.
$\therefore \angle ASB = 90^{\circ}$
$\Rightarrow BS \perp AS$
Since $BS$ is the radius of the circle,$AS$ must be a tangent to the circle. Similarly,$AR, BP,$ and $BQ$ are the tangents.

(N/A) Construction steps:
$1$. Construct $\triangle ABC$ with $AB = 6 \, cm$,$BC = 8 \, cm$ and $\angle B = 90^{\circ}$.
$2$. Draw $BD \perp AC$. Since $\angle BDC = 90^{\circ}$,the circle passing through $B, C, D$ will have $BC$ as its diameter because $\angle BDC = 90^{\circ}$ subtends a right angle at $D$.
$3$. Let $O$ be the midpoint of $BC$. Draw a circle with center $O$ and radius $OB = OC = 4 \, cm$. This circle passes through $B, C$ and $D$ (since $\angle BDC = 90^{\circ}$).
$4$. To construct tangents from $A$ to this circle,join $AO$. Bisect $AO$ at $M$. Taking $M$ as center and $MA$ as radius,draw a circle. Let this circle intersect the circle with center $O$ at $B$ and $E$. Join $AE$. Thus,$AB$ and $AE$ are the required tangents.
Justification:
Join $OE$. Since $AB$ is a tangent to the circle at $B$,$\angle ABO = 90^{\circ}$. Since $AE$ is a tangent to the circle at $E$,$\angle AEO = 90^{\circ}$. In the circle with center $O$,$OB$ and $OE$ are radii. Since $AB$ and $AE$ are tangents from an external point $A$,$AB = AE$.

(N/A) To construct tangents to a circle drawn with a bangle (where the centre is unknown):
$1$. Draw two non-parallel chords $BC$ and $CD$.
$2$. Draw the perpendicular bisectors of $BC$ and $CD$. The point where these bisectors intersect is the centre $E$ of the circle.
$3$. Take a point $A$ outside the circle. Join $AE$.
$4$. Bisect $AE$ at point $F$. With $F$ as the centre and $FA$ as the radius,draw a circle.
$5$. Let this circle intersect the original circle at points $B$ and $G$. Join $AB$ and $AG$.
$AB$ and $AG$ are the required tangents.
Justification:
Join $EB$ and $EG$. $\angle ABE$ is an angle in a semi-circle,so $\angle ABE = 90^{\circ}$. Since $EB$ is a radius,$AB$ must be a tangent. Similarly,$\angle AGE = 90^{\circ}$,so $AG$ is a tangent.

(N/A) $\triangle A'BC'$ whose sides are $\frac{3}{4}$ of the corresponding sides of $\triangle ABC$ can be drawn as follows:
$1.$ Draw a $\triangle ABC$ with side $BC = 6 \, cm$,$AB = 5 \, cm$ and $\angle ABC = 60^{\circ}$.
$2.$ Draw a ray $BX$ making an acute angle with $BC$ on the opposite side of vertex $A$.
$3.$ Locate $4$ points (as $4$ is the greater number in $\frac{3}{4}$),$B_1, B_2, B_3, B_4$,on line segment $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$4.$ Join $B_4C$ and draw a line through $B_3$ parallel to $B_4C$ intersecting $BC$ at $C'$.
$5.$ Draw a line through $C'$ parallel to $AC$ intersecting $AB$ at $A'$. $\triangle A'BC'$ is the required triangle.
Justification:
The construction can be justified by proving $A'B = \frac{3}{4} AB$,$BC' = \frac{3}{4} BC$,$A'C' = \frac{3}{4} AC$.
In $\triangle A'BC'$ and $\triangle ABC$:
$\angle A'C'B = \angle ACB$ (Corresponding angles)
$\angle A'BC' = \angle ABC$ (Common)
$\therefore \triangle A'BC' \sim \triangle ABC$ ($AA$ similarity criterion)
$\Rightarrow \frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC}$ ........$(1)$
In $\triangle BB_3C'$ and $\triangle BB_4C$:
$\angle B_3BC' = \angle B_4BC$ (Common)
$\angle BB_3C' = \angle BB_4C$ (Corresponding angles)
$\therefore \triangle BB_3C' \sim \triangle BB_4C$ ($AA$ similarity criterion)
$\Rightarrow \frac{BC'}{BC} = \frac{BB_3}{BB_4} = \frac{3}{4}$ ........$(2)$
From equations $(1)$ and $(2)$,we obtain:
$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{3}{4}$
$A'B = \frac{3}{4} AB, BC' = \frac{3}{4} BC, A'C' = \frac{3}{4} AC$.
This justifies the construction.