Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are $1\frac{1}{2}$ times the corresponding sides of the isosceles triangle.

detailed solution

1

• Draw the line segment of the base 8 cm. Draw perpendicular bisector of the line. Mark a point on the bisector which measures 4 cm from the base. Connect this point from both ends.
• Then draw another line that makes an acute angle with the given line. Divide the line into m + n parts where m and n are the ratios given.
• Two triangles are said to be similar if their corresponding angles are equal. They are said to satisfy Angle-Angle-Angle (AAA) Axiom.
• The basic proportionality theorem states that “If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally".

Steps of construction:

1. Draw BC = 8cm.
2. Through D, the mid-point of BC, draw the perpendicular to BC and cut an arc from D on it such that DA = 4cm. Join BA and CA. ΔABC is obtained.
3. Draw the ray BX so that ∠CBX is acute.
4. Mark 3 (since, 3 > 2 in 112
5. = 3/2) points B₁, B₂, B₃ on BX such that BB₁ = B₁B₂ = B₂B₃
6. Join B₂ (2^nd point ∵ 2 < 3) to C and draw B₃C' parallel to B₂C, intersecting BC extended at C’.
7. Through C’ draw C'A  to CA to intersect BA extended to A’. Now, ΔA'BC' is the required triangle similar to ΔABC where BA'/BA = C'A'/CA = BC'/BC = 3/2

Proof:

In ΔBB₃C', B₂C || B₃C',

Hence,

B₂B₃/BB₂ = CC'/BC = 1/2

Adding 1 to CC'/BC = 1/2

CC'/BC + 1 = 1/2 + 1

(BC+CC')/BC = 3/2

BC'/BC = 3/2

Consider ΔBAC and ΔBA'C'

∠ABC = ∠A'BC' (Common)

∠BCA = ∠BC'A' (Corresponding angles ∵ CA || C'A')

∠BAC = ∠BA'C' (Corresponding angles)

By AAA axiom, ΔBAC ~ ΔBA'C'

∴ Corresponding sides are proportional

Hence,

BA'/BA = BC'/BC= C'A'/CA = 3/2