## If the equal sides AB and AC of an isosceles triangle be produced to E and F SO that BE. CF = AB^2, show that the line EF will always pass through a fixed point.

How do I approach this problem

04-Oct-2015 11:29 PM
~6

10

AB = AC = t and Angle between AB and AC is pi/3

In pic A is at the origin.

Let,

BE = AB*x = t*w

CF = AC*y = t*v

Given, Ab^2=BE*CF  =>  t^2=(t*w)*(t*v)

Therefore, w*v=1 --------------------------------------eq(1)

Now let AB be x-axis then,

Coordinates of E( t*(1+w) , 0 )

Coordinates of F( t*(1+v)*cos(pi/3) , t*(1+v)*sin(pi/3) )

For simplicity assume a=sin(pi/3) and b=cos(pi/3) as they are constants

Equation of EF,

(y - 0) = (x - t*(1+w)) * a*(1+v) / ( b*(1+v) - (1+w) ) --------------------eq2

Substitue v from eq(1) in eq(2)

y = (x - t*(1+w)) * ( a/(b-w) ) ----------------------- General eq of EF

NOW, we have to prove that for any value of w(viz. w1 and w2) coordinates are independent of w1 and w2

y = (x - t*(1+w1)) * ( a/(b-w1) ) ----------------------- eq(3)

y = (x - t*(1+w2)) * ( a/(b-w2) ) ----------------------- eq(4)

Solving both of them we get,

x = b*t+t    and    y = a*t

Both are indepenndent of w1 and w2, HENCE PROVED!!

05-10-2015 01:27
~40

2

Try to draw a triangle randomly and then take AE'=AF and AF'=AE (E' and F' are on AB and AC respectively).

Draw a parallel line passing through intersection of EF and EF'. Prove congruency.

05-10-2015 01:14
~12

1

Since you asked for an approach I'm not giving a full solution.

Let P and Q be the points on AB and AC produced such that AB=BP and AC=CQ. Join EF and let it intersect PQ at R. Using whatever way you wish, find the ratio PR:RQ.

Spoiler: The ratio will come out a constant.

05-10-2015 01:12
~48