An Olympiad Geometry problem with incenter configurations.Let the inscribed circle of triangle ABC touches side BC at D ,side CA at E and side AB at F. Let G be the foot of the perpendicular from D to EF. Show that F G E G = B F C E .So this problem is equivalent to proving similarity between Δ B F G and Δ C E G.I was able to prove that ∠ G F B= ∠ G E C but after that, I hit a dead end. I found the point G pretty annoying as I couldn't apply any circle theorems to it.

Question

An Olympiad Geometry problem with incenter configurations.Let the inscribed circle of triangle ABC touches side BC at D ,side CA at E and side AB at F. Let G be the foot of the perpendicular from D to EF. Show that             F      G              E      G        =            B      F              C      E      .So this problem is equivalent to proving similarity between   Δ  B  F  G and   Δ  C  E  G.I was able to prove that   ∠  G  F  B=    ∠  G  E  C but after that, I hit a dead end. I found the point G pretty annoying as I couldn&#039;t apply any circle theorems to it.

vakleraarrc · Accepted Answer

Explanation:Just direct angle chasing and trigo will help you express those ratios.- Show that   B  F      /    C  E  =  B  D      /    D  C  =  (  B  D      /    D  G  )      /    (  D  C      /    D  G  )  =  tan  ⁡  (      90    ∘    −  β      /    2  )      /    tan  ⁡  (      90    ∘    −  γ      /    2  )  =  tan  ⁡  γ      /    2      /    tan  ⁡  β      /    2- Show that   ∠  I  F  E  =  ∠  I  A  E  =  α      /    2.- Show that   F  G      /    G  E  =  (  F  G      /    G  D  )      /    (  G  E      /    G  D  )  =  tan  ⁡  γ      /    2      /    tan  ⁡  β      /    2.

st3he1d0t · Accepted Answer

Step 1

Let the inscribed circle of triangle ABC touches side BC at D, side CA at E and side AB at F. Let G be the foot of the perpendicular from D to EF. Show that (FG)/(EG)=(BF)/(CE).

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