1. **Stating the problem:** We have a right triangle $\triangle ABC$ with $m(\angle B) = 90^\circ$, $m(\angle C) = 60^\circ$, and side $BC = 4$ cm. The bisector of angle $B$ intersects side $AC$ at $E$ and the line through $A$ parallel to $BC$ at $D$. We need to: (a) Show that the perimeter of $\triangle ABD$ is $4(2\sqrt{3} + \sqrt{6})$ cm. (b) Find the length of segment $BE$. 2. **Known facts and formulas:** - In a right triangle, the sum of angles is $180^\circ$. - The angle bisector theorem: the bisector divides the opposite side into segments proportional to the adjacent sides. - The perimeter is the sum of the lengths of all sides. - Parallel lines imply corresponding sides are equal in length. 3. **Find missing sides of $\triangle ABC$:** Since $m(\angle B) = 90^\circ$ and $m(\angle C) = 60^\circ$, then $m(\angle A) = 30^\circ$. Using the side $BC = 4$ cm opposite angle $A$ (30°), and knowing the triangle is right angled at $B$: - $BC$ is opposite $A$ (30°) - $AB$ is opposite $C$ (60°) - $AC$ is the hypotenuse Using the ratios in a 30°-60°-90° triangle: $$AB = BC \times \sqrt{3} = 4\sqrt{3}$$ $$AC = 2 \times BC = 8$$ 4. **Coordinates for clarity (optional):** Place $B$ at origin $(0,0)$, $C$ at $(4,0)$, and $A$ at $(0,4\sqrt{3})$. 5. **Find point $D$:** $D$ lies on the line through $A$ parallel to $BC$. Since $BC$ is horizontal, the line through $A$ parallel to $BC$ is horizontal at $y = 4\sqrt{3}$. $D$ is the intersection of this line with the extension of $AB$. The line $AB$ passes through $B(0,0)$ and $A(0,4\sqrt{3})$, so it is vertical $x=0$. Therefore, $D$ is at $(0,4\sqrt{3})$, which coincides with $A$. But the problem states $D$ is on the parallel through $A$ to $BC$, so $D$ must be on the extension of $AB$ or another line. Re-examining: The problem states $D$ is on the parallel through $A$ to $BC$ and on the bisector of angle $B$. 6. **Find the bisector of angle $B$:** Angle $B$ is $90^\circ$ at $B(0,0)$ between $BA$ and $BC$. - Vector $BA = A - B = (0,4\sqrt{3})$ - Vector $BC = C - B = (4,0)$ The bisector direction vector is the sum of the unit vectors along $BA$ and $BC$: - Unit vector $BA = \frac{(0,4\sqrt{3})}{|BA|} = (0,1)$ - Unit vector $BC = \frac{(4,0)}{4} = (1,0)$ Sum: $(1,1)$ Equation of bisector line from $B$: $$x = t, \quad y = t$$ 7. **Find $E$ as intersection of bisector and $AC$:** Line $AC$ passes through $A(0,4\sqrt{3})$ and $C(4,0)$. Slope of $AC$: $$m = \frac{0 - 4\sqrt{3}}{4 - 0} = -\sqrt{3}$$ Equation of $AC$: $$y - 4\sqrt{3} = -\sqrt{3}(x - 0) \Rightarrow y = 4\sqrt{3} - \sqrt{3}x$$ Set equal to bisector line $y = x$: $$x = 4\sqrt{3} - \sqrt{3}x$$ $$x + \sqrt{3}x = 4\sqrt{3}$$ $$x(1 + \sqrt{3}) = 4\sqrt{3}$$ $$x = \frac{4\sqrt{3}}{1 + \sqrt{3}}$$ Rationalize denominator: $$x = \frac{4\sqrt{3}(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{4\sqrt{3} - 12}{1 - 3} = \frac{4\sqrt{3} - 12}{-2} = 6 - 2\sqrt{3}$$ Then $y = x = 6 - 2\sqrt{3}$. So $E = (6 - 2\sqrt{3}, 6 - 2\sqrt{3})$. 8. **Find $D$ as intersection of bisector and line through $A$ parallel to $BC$:** Line through $A$ parallel to $BC$ is horizontal at $y = 4\sqrt{3}$. Set $y = 4\sqrt{3}$ in bisector line $y = x$: $$x = 4\sqrt{3}$$ So $D = (4\sqrt{3}, 4\sqrt{3})$. 9. **Calculate lengths for perimeter of $\triangle ABD$:** - $AB = 4\sqrt{3}$ (given) - $BD = $ distance between $B(0,0)$ and $D(4\sqrt{3},4\sqrt{3})$: $$BD = \sqrt{(4\sqrt{3})^2 + (4\sqrt{3})^2} = \sqrt{48 + 48} = \sqrt{96} = 4\sqrt{6}$$ - $AD = $ distance between $A(0,4\sqrt{3})$ and $D(4\sqrt{3},4\sqrt{3})$: $$AD = 4\sqrt{3}$$ 10. **Sum perimeter:** $$P_{ABD} = AB + BD + AD = 4\sqrt{3} + 4\sqrt{6} + 4\sqrt{3} = 8\sqrt{3} + 4\sqrt{6} = 4(2\sqrt{3} + \sqrt{6})$$ This proves part (a). 11. **Find length $BE$:** $B = (0,0)$, $E = (6 - 2\sqrt{3}, 6 - 2\sqrt{3})$ Calculate distance: $$BE = \sqrt{(6 - 2\sqrt{3})^2 + (6 - 2\sqrt{3})^2} = \sqrt{2(6 - 2\sqrt{3})^2}$$ Expand: $$(6 - 2\sqrt{3})^2 = 36 - 24\sqrt{3} + 12 = 48 - 24\sqrt{3}$$ So: $$BE = \sqrt{2(48 - 24\sqrt{3})} = \sqrt{96 - 48\sqrt{3}}$$ Factor out 48: $$BE = \sqrt{48(2 - \sqrt{3})} = 4\sqrt{3} \sqrt{2 - \sqrt{3}}$$ Simplify $\sqrt{2 - \sqrt{3}}$ by rationalizing: $$2 - \sqrt{3} = \left(\sqrt{3} - 1\right)^2$$ Check: $$(\sqrt{3} - 1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \neq 2 - \sqrt{3}$$ Try $(\sqrt{2} - \sqrt{1.5})^2$ or approximate: Approximate $2 - \sqrt{3} \approx 2 - 1.732 = 0.268$. So: $$BE \approx 4\sqrt{3} \times \sqrt{0.268} = 4\sqrt{3} \times 0.518 = 4 \times 1.732 \times 0.518 \approx 3.59$$ Exact form is $BE = \sqrt{96 - 48\sqrt{3}}$ cm. **Final answers:** - (a) $P_{ABD} = 4(2\sqrt{3} + \sqrt{6})$ cm - (b) $BE = \sqrt{96 - 48\sqrt{3}}$ cm