Let $P$ be a point on $DE$ and let the extensions of $BP$ and $CP$ meet the sides $AC$ and $AB$ at $G$ and $F$ respectively.

Question

In $\Delta ABC$ , $AB = AC = 1$ and $D,E$ are midpoints of $AB$ and $AC$ respectively. Let $P$ be a point on $DE$ and let the extensions of $BP$ and $CP$ meet the sides $AC$ and $AB$ at $G$ and $F$ respectively. Find $\frac{1}{BF} + \frac{1}{CG}$ .

What I Tried: Here is a picture :-

If we take $P$ at either ends like $D$ or $E$ then we get $BF = 0.5$ cm and $CG = 1$ cm . From here I get that $\frac{1}{BF} + \frac{1}{CG} = 3$ . But the surprising thing is wherever I choose $P$ the answer is always coming $3$ in Geogebra. I cannot understand how that is coming.

Can someone explain why the value is always constant? Thank You.

Answer 1

$\frac{GE}{GC} = \frac{PE}{BC}$ (by similar triangles $\triangle GPE, \triangle GBC$)

Similarly,

$\frac{FD}{FB} = \frac{PD}{BC}$

Adding both,

$\frac{GE}{GC} + \frac{FD}{FB} = \frac{1}{2}$

$\frac{GC-EC}{GC} + \frac{FB-BD}{FB} = \frac{1}{2}$

$2 - \frac{EC}{GC} - \frac{BD}{FB} = \frac{1}{2}$

As $EC = BD = \frac{1}{2}$

We get $\frac{1}{GC} + \frac{1}{FB} = 3$