In triangle ABC, D is the mid-point of BC. If DL perpendicular to AB and DM perpendicular to AC such that DL = DM, then the triangle will be:

Correct — D. DL and DM are the distances from D to the two sides, so compare areas.

D is the mid-point of BC, so BD = DC.
Equal bases with the same apex A give area(ABD) = area(ACD).

Measuring those same two areas from the other sides:
area(ABD) = ½ × AB × DL
area(ACD) = ½ × AC × DM

Equate them: ½ × AB × DL = ½ × AC × DM.
DL = DM cancels, leaving AB = AC.

Two equal sides makes it an isosceles triangle → option (d).

Why the others are wrong

• (a)
  right angled triangle — Nothing forces a 90° angle in ABC. The right angles here are at L and M, made by the construction of the perpendiculars, and they say nothing about angles A, B or C.
• (b)
  equilateral triangle — Equilateral is too strong. The condition yields AB = AC and leaves BC free, so an equilateral triangle is one special case that satisfies it rather than the conclusion it forces.
• (c)
  obtuse angle triangle — An obtuse angle is neither required nor excluded. The condition pins a relation between two sides, and an isosceles triangle may be acute, right or obtuse.

Two standard routes reach the same conclusion.

The area route: a median splits a triangle into two equal areas, and writing each of those areas against AB and AC with heights DL and DM forces AB = AC once DL = DM.

The congruence route: right triangles BLD and CMD match by RHS — right angles at L and M, hypotenuses BD = DC, legs DL = DM. That gives ∠B = ∠C, and equal base angles mean AB = AC.

Key facts

• A median divides a triangle into two triangles of equal area, because the bases are equal and the apex is shared.
• Right triangles BLD and CMD are congruent by RHS: equal hypotenuses BD = DC and equal legs DL = DM.
• Equal base angles ∠B = ∠C force the opposite sides AB and AC to be equal.
• The converse also holds: in an isosceles triangle the mid-point of the base is equidistant from the two equal sides.

Common traps

• Assuming AD bisects angle A and using that to prove AB = AC. The bisector property follows from AB = AC, so leaning on it first is circular.
• Jumping to equilateral because the described figure looks symmetric. Only two sides are pinned.
• Confusing DL = DM with AL = AM. The equal perpendiculars are dropped from D, not from A.

No directly related past PYQ was found.