Question

Triangle formed by joining the mid-points of the sides of an isosceles triangle is always a/an

• A. Scalene triangle
• B. Equilateral triangle
• C. Right-angled triangle
• D. Isosceles triangle

Answer 1

The correct option is D Isosceles triangle

Let ΔABC be an isosceles triangle such that AB = AC and D, E, F be the mid-points of sides AB, AC and BC respectively.
$\therefore BF=FC=\frac{1}{2}BC$
$AD=DB=\frac{1}{2}AB$
$AE=EC=\frac{1}{2}AC$
Also, AB = AC.
∴ AD = DB =AE = EC …..(i)
⇒ ∠ADE = ∠AED = ∠ABF = ∠ACF …..(ii) (By mid-point theorem, ∠ADE = ∠ABF and ∠AED = ∠ACF)
In ΔDBF and ΔECF,
DB = EC [From (i)]
∠DBF = ∠ECF [From (i)]
BF = CF (Given)
∴ ΔDBF ≅ ΔECF (SAS congruence rule)
⇒ DF = EF (CPCT)
In ΔADE, AD = AE therefore, it is an isosceles triangle whereas ΔDBF and ΔECF are not necessarily isosceles triangles.
Thus, ΔADE is not congruent with any of the triangles.
∴ DF = EF ≠ DE
⇒ ΔDEF is an isosceles triangle
Hence, the correct answer is option (d).