In ΔABC,
∠A + ∠B + ∠C = 180° (the sum of the interior angles of a Δ is equal to 180°)
35° + 100° + ∠C= 180°
⇒ ∠C = 180° – 135°
⇒ ∠C = 45°
But in ΔABD, BD ⊥ AC
i.e., ∠ADB = 90°
and ∠A = 35° (given)
∠ABD = 180° – 125° = 55°
∠DBC = 100° – 55° = 45°
Now, ∠DCB = ∠DBC = 45° (Proved above)
Hence BDC is an isosceles triangle.