Question

Three statements are given below:
I. In a ∆ABC in which AB = AC, the altitude AD bisects BC.
II. If the altitudes AD, BE and CF of ∆ABC are equal, then ∆ABC is equilateral.
III. If D is the midpoint of the hypotenuse AC of a right ∆ABC, then BD = AC.
Which is true?
(a) I only
(b) II only
(c) I and II
(d) II and III

Answer 1

(c) I and II are true
(I)

$$\begin{gathered} \mathrm{In}△\mathrm{ADB}\mathrm{and}△\mathrm{ADC},\mathrm{we}\mathrm{get}: \\ \mathrm{AB}=\mathrm{AC}\left(\mathrm{Given}\right) \\ \mathrm{AD}=\mathrm{AD}\left(\mathrm{Common}\right) \\ \angle \mathrm{ADB}=\angle \mathrm{ADC}=90°\left(\mathrm{Given}\right) \\ △\mathrm{ADB}\cong △\mathrm{ADC}(\mathrm{RHS}\hspace{0.17em}\mathrm{criterion}) \\ \Rightarrow \mathrm{BD}=\mathrm{DC}\left(\mathrm{CPCT}\right) \end{gathered}$$

∴ AD bisects BC, which is true.

(II)

$$\begin{gathered} \mathrm{In}△\mathrm{ABE}\mathrm{and}△\mathrm{ACF},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{AEB}=\angle \mathrm{AFC}(\mathrm{Each}90°) \\ \angle \mathrm{BAE}=\angle \mathrm{CAF}\left(\mathrm{Common}\right) \\ \mathrm{BE}=\mathrm{CF}\left(\mathrm{Given}\right) \\ ∆\mathrm{ABL}\cong △\mathrm{ACM}(\mathrm{AAS}\mathrm{criterion}) \\ \Rightarrow \mathrm{AB}=\mathrm{AC}\left(\mathrm{CPCT}\right) \end{gathered}$$
Similarly, if AD is perpendicular to BC and AD = BE, then we can prove that BC = AC.
Therefore, triangle ABC is an equilateral triangle.
Hence, (II) is true.

(III)

$\mathrm{In}△\mathrm{ABC},\angle \mathrm{ABC}=90°(\mathrm{BD}\mathrm{perpendicular}\mathrm{AC})$
Then we know that AD = BD = CD
Therefore, BD = AC is not true.
So, only (I) and (II) are true.