Discussion

In \(\triangle ABC\), \(\angle A = 90°\). The bisector of \(\angle C\) meets AB at D. If \(\angle B = 40°\), find \(\angle BDC\).

Answer: A

Correct answer: 115°

The angles of \(\triangle ABC\) sum to \(180°\), so

\[\angle C = 180° - 90° - 40° = 50°\]

CD bisects it, giving \(\angle ACD = 25°\).

In \(\triangle ADC\) the angles at A and C are \(90°\) and \(25°\), so \(\angle ADC = 65°\). Since D lies on AB, \(\angle BDC\) is its supplement:

\[\angle BDC = 180° - 65° = 115°\]

The exterior-angle theorem gives the same result directly: in \(\triangle ADC\), \(\angle BDC = \angle A + \angle ACD = 90° + 25° = 115°\).

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