Look at the statements given below: I. If AD, BE and CF be the altitudes of a ΔABC such that AD = BE = CF, then ΔABC is an equilateral triangle. II. If D is the mid-point of hypotenuse AC of a right ΔABC, then BD = AC. III. In an isosceles ΔABC 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

Question

Philoid · Accepted Answer

We can clearly observe that statement I and statement III are correct.

We can prove the statement as follows:

In ∆ABC, altitudes AD, BE and CF are equal

Now, In ∆ABE and ∆ACF,

BE = CF (Given)

∠A = ∠A (common)

∠AEB = ∠AFC (Each 90°)

Therefore, by AAS axiom,

∆ABE ≅ ∆ ACF

AB = AC (by cpct)

In the same way, ∆BCF ≅ ∆BAD

thus, BC = AB (by cpct)

Therefore AB = AC = BC

Thus, ∆ABC is an equilateral triangle.

We can prove the IIIrd statement as follows:

Let ∆ABC be an isosceles triangle with AD as an altitude

Now, In ∆ABD and ∆ADC,

AB = AC (Given)

∠B = ∠C (Angles opposite to equal sides)

∠BDA = ∠CDA (each 90°)

Therefore by AAS axiom,

∆ABD ≅ ∆ADC

BD = DC (by congruent parts of congruent triangles)

∴ D is the mid-point of BC and hence AD bisects BC.