An equilateral triangle $ABC$ has a circle inscribed within it. The radius of the circle is 4 cm. (i) Find the length of $OB$. (ii) Determine if $DE$ is parallel to $CA$. Provide a reason. (iii) (a) Find all the angles of quadrilateral $OEBD$ and show it is a cyclic quadrilateral. (iii) (b) Find the perimeter of triangle $ABC$. (Use $\sqrt{3} = 1.73$).

1. Problem Description

An equilateral triangle $ABC$ has a circle inscribed within it. The radius of the circle is 4 cm.

(i) Find the length of $OB$.

(ii) Determine if $DE$ is parallel to $CA$. Provide a reason.

(iii) (a) Find all the angles of quadrilateral $OEBD$ and show it is a cyclic quadrilateral.

(iii) (b) Find the perimeter of triangle $ABC$. (Use $\sqrt{3} = 1.73$).

2. Solution Steps

(i) Determine the length $OB$.

In an equilateral triangle, the center of the inscribed circle is the intersection of the angle bisectors. Therefore, $OB$ bisects angle $B$. Since $\angle ABC = 60^\circ$, then $\angle OBD = \frac{1}{2} \times 60^\circ = 30^\circ$.

$OD$ is the radius of the circle, so $OD = 4$ cm. Triangle $ODB$ is a right-angled triangle with $\angle ODB = 90^\circ$.

We can use trigonometry to find $OB$.

$\sin(\angle OBD) = \frac{OD}{OB}$

$\sin(30^\circ) = \frac{4}{OB}$

$OB = \frac{4}{\sin(30^\circ)} = \frac{4}{1/2} = 8$ cm.

(ii) Is $DE || CA$? Give reason.

Since $\angle ABC = 60^\circ$ and $\angle EBO = 30^\circ$, then $\angle OBE = 30^\circ$. Also, $\angle BEO = 90^\circ$ because the radius is perpendicular to the tangent at the point of contact. Therefore $\angle BEO = 90^\circ$.

In triangle $OBE$, $\angle BOE = 180^\circ - 90^\circ - 30^\circ = 60^\circ$.

$\angle BEA = 90^\circ$ so the line segment $BE$ is tangent to the circle at $E$.

In triangle $ABC$, since the triangle is equilateral, each angle is $60^\circ$. Therefore $\angle BAC = 60^\circ$. Since $\angle BEO = 90^\circ$, $BE$ is not parallel to $AC$. Also $\angle BED = 180 - 90 = 90$

If $DE || CA$, then $\angle EDB = \angle BCA = 60^\circ$. However, $\angle EDB = 90^\circ$, so $DE$ is not parallel to $CA$.

(iii)(a) Write all angles of quadrilateral $OEBD$ and show that it is a cyclic quadrilateral.

$\angle OEB = 90^\circ$ (tangent is perpendicular to radius)

$\angle OBD = 30^\circ$ (OB bisects angle B which is 60 degrees)

$\angle BDO = 90^\circ$ (tangent is perpendicular to radius)

$\angle EOD = 360^\circ - 90^\circ - 30^\circ - 90^\circ = 150^\circ$.

In quadrilateral OEBD, $\angle OEB + \angle ODB = 90^\circ + 90^\circ = 180^\circ$.

Since the sum of opposite angles is $180^\circ$, quadrilateral $OEBD$ is a cyclic quadrilateral.

(iii)(b) Find the perimeter of triangle $ABC$. (Use $\sqrt{3} = 1.73$).

The radius of the inscribed circle in an equilateral triangle is $r = \frac{a}{2\sqrt{3}}$ where $a$ is the side length of the triangle.

Given $r=4$, we have $4 = \frac{a}{2\sqrt{3}}$.

$a = 8\sqrt{3} = 8 \times 1.73 = 13.84$ cm.

The perimeter of the triangle is $3a = 3 \times 13.84 = 41.52$ cm.

3. Final Answer

(i) $OB = 8$ cm

(ii) No, $DE$ is not parallel to $CA$.

(iii) (a) $\angle OEB = 90^\circ$, $\angle OBD = 30^\circ$, $\angle BDO = 90^\circ$, $\angle EOD = 150^\circ$. $OEBD$ is a cyclic quadrilateral.

(iii) (b) Perimeter of triangle $ABC = 41.52$ cm.