Q5 of 23 Page 1

Solve any two sub-questions:

In the following figure, AE = EF = AF = BE = CF = a, AT ⊥ BC. Show that AB = AC = √3 x a.


The figure is as follows:


Given:


AT ⊥ BC


AE = EB = AF = RN = FC = a


To Prove: = AB = BC = √3 a


Proof:


In ABF,


EB = EF = a [Given]


∴ E is the midpoint of BF …(i)


∴ AE is a median. [From (i)]


∴ AB + AF2 = 2AE2 + 2EB2 [By Apollonius Theorem]


AB2 + a2 = 2a2 + 2a2


AB2 + a2 = 4a2


AB2 = 4a2- a2


AB2 = 3a2


AB = √3 a


AC = √3 a.


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