Question

Question 2
State whether the following is true or false.
To construct a triangle similar to a given $\Delta$ABC with its sides of $\frac{7}{3}$ corresponding sides of $\Delta$ABC, draw a ray BX making acute angle with BC and X lies on the opposite side of A with respect of BC. The points $B_1,B_2,...B_7$ are located at equal distances on BX, $B_3$ joined to C and then a line segment $B_{6}C$’ is drawn parallel to $B_{3}C$ where C’ lies on BC produced . Finally line segment A’C’ is drawn parallel to AC.

Answer 1

False
Steps of construction
Draw a line segment BC with suitable length
Taking B and C as centers draw two arcs of suitable radii intersecting each other at A
Join BA and CA. $\Delta$ ABC is the required triangle.
From B draw any ray BX downwards making an acute angle CBX.
Locate seven points $B_1,B_2,B_3....B_7$ on BX such that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4=B_4{B}_5=B_5{B}_6=B_6{B}_7$.
Join $B_{3}C$ and form $B_7$ draw a line $B_{7}C$’ || $B_{3}C$ intersecting the extended line segment BC at C’.
From point C’ draw C’A || CA intersecting the extended line segment BA at A’. then $\Delta$ A’BC’ is the required triangle whose sides are of the corresponding sides of $\Delta$ABC. Given that, segment $B_{6}C$’ is drawn parallel to $B_{3}C$. But form our construction is never possible that segment $B_{6}C$’ is parallel to $B_{3}C$ because the similar triangle A’BC’ has its sides of the corresponding sides of triangle ABC. So . $B_{7}C$’ is parallel to $B_{3}C$.