Question

Look at the statements given below:
I. If AD, BE and CF be the altitudes of a ∆ABC such that AD = BE = CF, then ∆ABC is an equilateral triangle.
II. IF D is the mid-point of hypotenuse AC of a right ∆ABC, then BD = AC.
III. In an isosceles ∆ABC in which AB = AC, the altitude AD bisects BC.
Which is true?
(a) I only
(b) II only
(c) I and III
(d) II and III

Answer 1

(c) I and III

Explanation:

Statements I and III are true.
For I:
In ​∆ABC, the altitudes AD = BE = CF .
In ∆ABE and ∆ACF, we have:
BE = CF (Given)
∠A is common.
∠​AEB = ∠​AFC = 90^o
i.e., ∆ABE ≅​ ∆ ACF ( By AAS congruence rule)
∴ AB = AC (CPCT)
Similarly, ∆BCF ≅​ ∆BAD
i.e., BC = AB ( CPCT)
∴AB = AC = BC

Hence, ∆ABC is an equilateral triangle.

For III:

Let ​∆ABC be an isosceles triangle. AD is the altitude of ​∆ABC.
In ​ ​∆ABD and ​∆ADC, we have:
AB = AC (Given)
∠B = ​​​∠C (Angles opposite to equal sides are equal)
∠ADB = ​∠ADC = 90^o
So, ∆ABD ≅ ​​∆ADC (By AAS conruency)
⇒ BD = DC
∴ D is the mid point of BC or AD bisects BC.