In △ABC, a median AD is drawn to side BC. What is the ratio of the area of △ABD to the area of △ACD?

Correct — A.
A median joins a vertex to the midpoint of the opposite side, so BD = DC
△ABD and △ACD share apex A and have bases on line BC, so they share one height h
Area ratio = (½ × BD × h) : (½ × DC × h) = BD : DC

BD = DC gives 1 : 1 → option (a)

Why the others are wrong

• (b)
  1:2 — 1 : 2 would need DC = 2 × BD, which puts D one-third of the way from B to C. A median ends at the midpoint, not a trisection point.
• (c)
  2:1 — 2 : 1 is the centroid's ratio: G divides median AD so that AG : GD = 2 : 1. That is a ratio of lengths along the median, not of these two areas.
• (d)
  1:3 — 1 : 3 would need DC = 3 × BD, putting D a quarter of the way from B to C. A median lands at the midpoint, so BD and DC are equal.

Triangles with the same height have areas in the ratio of their bases. △ABD and △ACD both hang from vertex A over the line BC, so their heights are equal and their areas follow BD : DC.

A median makes BD = DC, so it cuts the triangle into two equal areas. Each half is ½ of △ABC, however lopsided the triangle is.

Key facts

• A median divides a triangle into two triangles of equal area
• Triangles with equal heights have areas in the ratio of their bases
• The medians meet at the centroid, which divides each in the ratio 2 : 1 from the vertex
• The three medians cut a triangle into six small triangles of equal area

Common traps

• Carrying the centroid's 2 : 1 over to the areas. That ratio splits the length of the median, not △ABC.
• Expecting the answer to depend on the triangle's shape. Equal bases and a shared height give 1 : 1 for any triangle.

No directly related past PYQ was found.