In an equilateral triangle ABC, AD is altitude drawn from A on side BC. prove that (AB/AD)² = 4/3 .?

In right triangle ADB and right triangle ADC,

(i)  AB = AC  (given)
(ii) AD = AD (common)

Therefore the two triangles are congruent.

=>  BD  =  CD  ................ (1)

Now,  BC  =  BD + CD  =  BD + BD ...... [ using (1) above) ]

=>  BC  =  2 BD  ====>>>  BD  =  BC/2  ................... (2)

In triangle ABD, we have

AB²  =  AD² + BD²  ............... ( by Pythagoras theorem )

=>  AB²  =  AD² + (BC/2)²  ............. Using (2)

=>  AB²  =  AD² +  BC²/4

=>  4 AB²  =  4 AD² +  BC²  =  4 AD² +  AB²  .....  ( since BC = AB )

=> 4 AB² - AB²  =  4 AD²

=> 3 AB²  =  4 AD²

=> ( AB/AD )²  =  4/3  ..............Proved

Refer to the figure below.

Let 2k be the length of each side of the equilateral ΔABC.

Since the altitude of equilateral triangle bisects the base,
then BD = DC = k

In ΔABC:
AB² = AD² + BD² (Pythagorean theorem)
(2k)² = AD² + k²
4k² = AD² + k²
AD² = 3k²

(AB/AD)² = AB²/AD²
(AB/AD)² = (2k)²/(3k²)
(AB/AD)² = (4k²)/(3k²)
(AB/AD)² = 4/3

In the triangle, "AD" is altitude.
AD²+DB²=AB²AD²+(1/2 * BC)²=AB²AD²+BC²/4=AB²4AD²+BC²=4AB²4AD²=4AB²-BC²4AD²=4AB²-AB²4AD²=3AB²AD²=3/4*AB²
(AB/AD)² = 4/3


Review it!

Suppose ABC's an equilateral triangle with side length = s. Draw a perpendicular from A to mid-
point of BC at D. Then triangle ABD is a 30-60-90 degree triangle with AB = 2s, BD = (1/2)s and
AD = srt3. Then (AB/AD)^2 = (2s/srt3)^2 = (2/rt3)^2 = (4/3).