In a kite ABCD, longer diagonal BD is drawn. If ∠ ABD = 54°and ∠ ADB = 50°, then what is the measure (in degrees) of the ∠ DCB?

Correct — D. In a kite the longer diagonal is the axis of symmetry, so BD reflects one half of the figure onto the other.

That reflection carries A to C and fixes B and D, so triangle ABD maps onto triangle CBD:
∠CBD = ∠ABD = 54°
∠CDB = ∠ADB = 50°

In triangle BCD the three angles sum to 180°:
∠DCB = 180 − 54 − 50 = 76° → option (d)

Why the others are wrong

• (a)
  108 — 108° is 2 × 54, the two halves at B added into the full angle ∠ABC. That is the kite's angle at B, not the angle at C that the question asks for.
• (b)
  100 — 100° is 2 × 50, the full angle ∠ADC at D. Same doubling as option (a), applied at the other end of the diagonal.
• (c)
  54 — 54° is ∠ABD copied straight from the stem. It lives at vertex B in triangle ABD, while ∠DCB is the third angle of the mirrored triangle BCD.

A kite has two pairs of equal adjacent sides. In ABCD that means AB = CB and AD = CD, so both B and D are equidistant from A and from C — they lie on the perpendicular bisector of AC.

Hence BD is the axis of symmetry, and it is the longer diagonal, exactly as the stem says. Reflecting in BD gives triangle ABD ≅ triangle CBD, so every angle in one has an equal twin in the other.

That hands you ∠CBD = 54° and ∠CDB = 50° for free, and the angle sum of triangle BCD finishes the job.

Key facts

• In a kite, the diagonal joining the vertices where the equal sides meet is the axis of symmetry.
• That axis bisects the other diagonal at right angles.
• Symmetry gives ∠ABD = ∠CBD and ∠ADB = ∠CDB in kite ABCD.
• It also makes the remaining pair of angles equal, so ∠BAD = ∠BCD = 76° here.

Common traps

• Assuming both diagonals bisect each other as in a parallelogram, when in a kite only the axis bisects the other diagonal.
• Doubling 54 or 50 and answering with the angle at B or at D.
• Reading ∠DCB as an angle at D, when the middle letter names the vertex.

No directly related past PYQ was found.