Q.From vertex A of triangle \(\triangle ABC\), a perpendicular AD is drawn to the base BC, intersecting BC at point D. If \(AD^2 = BD \cdot CD\), prove that triangle ABC is a right-angled triangle and that \(\angle A = 90^\circ\).

Between \(\triangle BDA\) and \(\triangle ADC\):
\(\angle BDA = \angle ADC = 90^\circ\) [\(\because\) AD \(\perp\) BC]
And, \(AD^2 = BD \cdot DC\)
That is, \(\frac{BD}{AD} = \frac{AD}{DC}\)
\(\therefore \triangle BDA\) and \(\triangle ADC\) are similar.

[Since, if one angle of a triangle is equal to one angle of another triangle, and the sides including those angles are proportional, then the triangles are similar.]

Therefore, \(\angle CAD = \angle ABD\) and \(\angle BAD = \angle ACD\)
\(\therefore \angle CAD + \angle BAD = \angle ABD + \angle ACD\)
or, \(\angle BAC = \angle ABD + \angle ACD\)
or, \(\angle BAC + \angle BAC = \angle ABD + \angle ACD + \angle BAC\)
or, \(2\angle BAC = 180^\circ\)
\(\therefore \angle BAC = 90^\circ\)
That is, \(\angle A = 90^\circ\)

Hence, \(\triangle ABC\) is a right-angled triangle with \(\angle A = 90^\circ\).
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