Question

1. if g is the incenter of $\triangle abc$, find each missing measure.

a) $gd = $ ______
b) $bg = $ ______
c) $fc = $ ______
d) $bf = $ ______

Explanation:

Step1: Recall Incenter Properties

The incenter \( G \) of a triangle is equidistant from all sides, so \( GE = GD = GF \), and the tangents from a vertex to the incircle are equal, so \( AE = AD \), \( BE = BF \), \( CF = CD \).

Step2: Find \( GD \) (Part a)

We know \( BE = 6 \) (from \( BE \) length), and \( GE = GD \) (incenter equidistant to sides). Also, in right triangle \( CGF \), \( CG = 37 \), \( CD = 35 \). Wait, first, for \( GD \): since \( GE = GD \) (incenter to sides \( AB \) and \( AC \) are equal), but also, in right triangle \( BEG \), \( BE = 6 \), and maybe we can find \( GD \) using the fact that \( GD \) is equal to \( GE \), but also, let's check \( CD = 35 \), \( CG = 37 \), so in right triangle \( CGD \), \( GD=\sqrt{CG^{2}-CD^{2}}=\sqrt{37^{2}-35^{2}}=\sqrt{(37 - 35)(37 + 35)}=\sqrt{2\times72}=\sqrt{144}=12 \). So \( GD = 12 \).

Step3: Find \( BG \) (Part b)

In right triangle \( BEG \), \( BE = 6 \), \( GE = GD = 12 \), so \( BG=\sqrt{BE^{2}+GE^{2}}=\sqrt{6^{2}+12^{2}}=\sqrt{36 + 144}=\sqrt{180}=6\sqrt{5}\)? Wait, no, wait \( BE = 6 \), \( GE = 12 \)? Wait, no, earlier \( GD = 12 \), so \( GE = GD = 12 \)? Wait, \( BE \) is 6, so in right triangle \( BEG \), \( BE = 6 \), \( GE = 12 \), then \( BG=\sqrt{6^{2}+12^{2}}=\sqrt{36 + 144}=\sqrt{180}=6\sqrt{5}\)? Wait, no, maybe I made a mistake. Wait, \( BE = 6 \), \( GE = GD = 12 \)? Wait, let's recalculate \( GD \): \( CG = 37 \), \( CD = 35 \), so \( GD=\sqrt{37^{2}-35^{2}}=\sqrt{1369 - 1225}=\sqrt{144}=12 \). Correct. Then \( BE = 6 \), \( GE = 12 \), so \( BG=\sqrt{6^{2}+12^{2}}=\sqrt{36 + 144}=\sqrt{180}=6\sqrt{5}\)? No, wait \( 6^{2}+12^{2}=36 + 144 = 180=36\times5 \), so \( \sqrt{180}=6\sqrt{5}\approx13.42 \)? Wait, no, maybe \( BE = 6 \), \( GE = 12 \), but maybe \( BE \) is 6, and \( GE = GD = 12 \), so \( BG=\sqrt{6^{2}+12^{2}} = \sqrt{36 + 144}=\sqrt{180}=6\sqrt{5}\)? Wait, but maybe I messed up \( BE \). Wait, the length \( BE \) is 6, as given (from the diagram, \( BE = 6 \)). So \( BG=\sqrt{6^{2}+12^{2}} = 6\sqrt{5}\)? Wait, no, 6 squared is 36, 12 squared is 144, sum is 180, square root of 180 is 6*sqrt(5) ≈13.42? Wait, no, 13.42 is not an integer. Wait, maybe I made a mistake in \( GD \). Wait, \( CG = 37 \), \( CD = 35 \), so \( GD=\sqrt{37^{2}-35^{2}}=\sqrt{(37 - 35)(37 + 35)}=\sqrt{2\times72}=\sqrt{144}=12 \). That's correct. Then \( BE = 6 \), \( GE = 12 \), so \( BG=\sqrt{6^{2}+12^{2}} = \sqrt{36 + 144}=\sqrt{180}=6\sqrt{5}\)? Wait, maybe the diagram has \( BE = 6 \), \( AE = AD \), \( CD = 35 \), \( CF = CD = 35 \)? Wait, no, part c is \( FC \), so \( FC = CD \) because tangents from \( C \) to the incircle are equal, so \( FC = CD = 35 \)? Wait, no, \( CD = 35 \), so \( FC = 35 \)? Wait, but in the diagram, \( CF \) is adjacent to \( CG = 37 \), \( FG = GD = 12 \), so \( FC=\sqrt{CG^{2}-FG^{2}}=\sqrt{37^{2}-12^{2}}=\sqrt{1369 - 144}=\sqrt{1225}=35 \). Ah! There we go. So \( FC = 35 \), which is equal to \( CD \), as tangents from \( C \) are equal. So that's part c. Then part d: \( BF = BE = 6 \), because tangents from \( B \) to the incircle are equal, so \( BE = BF = 6 \). Wait, let's re - do:

Correcting Step2 (GD):

The incenter \( G \) is equidistant from all sides, so \( GE = GD = GF \). In right triangle \( CGD \), \( CG = 37 \), \( CD = 35 \). By Pythagoras: \( GD=\sqrt{CG^{2}-CD^{2}}=\sqrt{37^{2}-35^{2}}=\sqrt{(37 - 35)(37 + 35)}=\sqrt{2\times72}=\sqrt{144}=12 \). So \( GD = 12 \) (part a).

Step for BG (part b):

In right triangle \( BEG \), \( BE = 6 \), \( GE = GD = 12 \). By Pythagoras: \( BG=\sqrt{BE^{2}+GE^{2}}=\sqrt{6^{2}+12^{2}}=\sqrt{36 + 144}=\sqrt{180}=6\sqrt{5}\)? Wait, no, 6 squared is 36, 12 squared is 144, sum is 180, square root of 180 is \( 6\sqrt{5}\approx13.42 \), but that seems odd. Wait, maybe \( BE \) is not 6? Wait, the diagram shows \( BE = 6 \), \( AE \) is some length, \( AD = AE \), \( CD = 35 \), \( FC = 35 \) (since \( FC = CD \) as tangents from \( C \)), \( BF = BE = 6 \) (tangents from \( B \)). Then for \( BG \): in right triangle \( BEG \), \( BE = 6 \), \( GE = 12 \), so \( BG=\sqrt{6^{2}+12^{2}}=\sqrt{36 + 144}=\sqrt{180}=6\sqrt{5}\)? No, wait \( 6^{2}+12^{2}=36 + 144 = 180 \), \( \sqrt{180}=6\sqrt{5}\approx13.42 \), but maybe I made a mistake in \( GD \). Wait, no, \( CG = 37 \), \( CD = 35 \), so \( GD = 12 \) is correct.

Step for FC (part c):

Since \( G \) is the incenter, \( FC = CD \) (tangents from point \( C \) to the incircle are equal). \( CD = 35 \), so \( FC = 35 \).

Step for BF (part d):

Since \( G \) is the incenter, \( BF = BE \) (tangents from point \( B \) to the incircle are equal). \( BE = 6 \), so \( BF = 6 \).

Re - checking BG:

Wait, maybe \( BE = 6 \), \( GE = 12 \), so \( BG=\sqrt{6^{2}+12^{2}}=\sqrt{36 + 144}=\sqrt{180}=6\sqrt{5}\approx13.42 \), but maybe the problem expects an integer. Wait, maybe I misread \( BE \). Wait, the diagram has \( BE = 6 \), \( AE \) is not given, \( AD = AE \), \( CD = 35 \), \( CG = 37 \), \( FG = 12 \), \( FC = 35 \), \( BF = 6 \), \( GD = 12 \), \( BG=\sqrt{6^{2}+12^{2}} = 6\sqrt{5}\)? No, 6 and 12, the right triangle with legs 6 and 12, hypotenuse is \( \sqrt{36 + 144}=\sqrt{180}=6\sqrt{5}\approx13.42 \), but maybe the problem has \( BE = 8 \)? No, the diagram says 6. Wait, maybe I made a mistake in \( GD \). Wait, \( CG = 37 \), \( CD = 35 \), so \( GD=\sqrt{37^{2}-35^{2}} = 12 \), that's correct. Then \( GE = 12 \), \( BE = 6 \), so \( BG=\sqrt{6^{2}+12^{2}}=6\sqrt{5}\). But let's proceed with the correct tangent properties:

- Tangents from a vertex to the incircle are equal: \( AE = AD \), \( BE = BF \), \( CF = CD \).
- Incenter is equidistant from sides: \( GE = GD = GF \).

So:

a) \( GD \): In \( \triangle CGD \), right - angled at \( D \), \( GD=\sqrt{CG^{2}-CD^{2}}=\sqrt{37^{2}-35^{2}} = 12 \).

b) \( BG \): In \( \triangle BEG \), right - angled at \( E \), \( BE = 6 \), \( GE = GD = 12 \), so \( BG=\sqrt{BE^{2}+GE^{2}}=\sqrt{6^{2}+12^{2}}=\sqrt{36 + 144}=\sqrt{180}=6\sqrt{5}\approx13.42 \)? Wait, no, 6 and 12, 6 - 12 - 6√5 is correct. But maybe the problem has \( BE = 8 \)? No, the diagram shows 6.

c) \( FC \): Since \( CF = CD \) (tangents from \( C \)), and \( CD = 35 \), so \( FC = 35 \).

d) \( BF \): Since \( BF = BE \) (tangents from \( B \)), and \( BE = 6 \), so \( BF = 6 \).

Answers:

a) \( GD=\boldsymbol{12} \)

b) \( BG=\boldsymbol{6\sqrt{5}} \) (or approximately \( 13.42 \))

c) \( FC=\boldsymbol{35} \)

d) \( BF=\boldsymbol{6} \)

Answer:

s:
a) \( GD=\boldsymbol{12} \)

b) \( BG=\boldsymbol{6\sqrt{5}} \) (or approximately \( 13.42 \))

c) \( FC=\boldsymbol{35} \)

d) \( BF=\boldsymbol{6} \)