图片挂了
(1)
第一问还是比较简单的，所以简写一下。
∵ ∠ E D G = 6 0 ∘ ∴ ∠ E D B + ∠ B E D = ∠ E D B + ∠ C D G = 12 0 ∘ ∴ ∠ B E D = ∠ C D G ∴ △ B E D ∼ △ C D G \because \angle EDG=60^\circ
\\
\therefore \angle EDB+\angle BED= \angle EDB+\angle CDG=120 ^\circ
\\
\therefore \angle BED=\angle CDG
\\
\therefore \bigtriangleup BED \sim \bigtriangleup CDG
\\ ∵ ∠ E D G = 6 0 ∘ ∴ ∠ E D B + ∠ BE D = ∠ E D B + ∠ C D G = 12 0 ∘ ∴ ∠ BE D = ∠ C D G ∴ △ BE D ∼ △ C D G (2)
过程有点多，写个大纲差不多了。
这题让我们判断 C △ A E G C_{\bigtriangleup AEG} C △ A EG △ A E G \bigtriangleup AEG △ A EG E G EG EG A G AG A G E E E A C AC A C
图片挂了
这样是不是就把 E G EG EG A G AG A G E H EH E H A G = A H − G H AG=AH-GH A G = A H − G H A E AE A E A H AH A H E H EH E H B E = a ， A E = 6 − a BE=a，AE=6-a BE = a ， A E = 6 − a A G AG A G a a a C G CG CG C G CG CG B E BE BE
有了上面的准备之后，剩下的直接爆算。
∵ △ B E D ∼ △ C D G ∴ B D C G = B E C D ⇒ 3 C G = a 3 ⇒ C G = 9 a ∴ A G = A C − C G = 6 − 9 a \because \bigtriangleup BED \sim \bigtriangleup CDG
\\
\therefore \frac{BD}{CG}=\frac{BE}{CD} \Rightarrow \frac{3}{CG}=\frac{a}{3} \Rightarrow CG=\frac{9}{a}
\\
\therefore AG=AC-CG=6-\frac{9}{a} ∵ △ BE D ∼ △ C D G ∴ CG B D = C D BE ⇒ CG 3 = 3 a ⇒ CG = a 9 ∴ A G = A C − CG = 6 − a 9 至此，△ A E G \bigtriangleup AEG △ A EG A E AE A E A G AG A G a a a E G EG EG E G EG EG ∠ A H E = 9 0 ∘ ， A E = a \angle AHE=90^\circ，AE=a ∠ A H E = 9 0 ∘ ， A E = a A H AH A H E H EH E H a a a
E H = A E × sin A = 3 3 − 3 2 a A H = A E × cos A = 1 2 × ( 6 − a ) = 3 − a 2 G H = G C − C H = ( A C − A G ) − ( A C − A H ) = A C − A G − A C + A H = A H − A G = 3 − a 2 − 6 − 9 a = 9 a − 3 − a 2 EH=AE \times \sin A=3 \sqrt{3} - \frac{\sqrt{3}}{2}a
\\
AH=AE \times \cos A=\frac{1}{2} \times (6-a)=3-\frac{a}{2}
\\
GH=GC-CH=(AC-AG)-(AC-AH)=AC-AG-AC+AH=AH-AG=3-\frac{a}{2} - 6-\frac{9}{a}=\frac{9}{a}-3-\frac{a}{2} E H = A E × sin A = 3 3 − 2 3 a A H = A E × cos A = 2 1 × ( 6 − a ) = 3 − 2 a G H = GC − C H = ( A C − A G ) − ( A C − A H ) = A C − A G − A C + A H = A H − A G = 3 − 2 a − 6 − a 9 = a 9 − 3 − 2 a 然后在 △ E G H \bigtriangleup EGH △ EG H
E G = G H 2 + E H 2 = a − 3 + 9 a EG=\sqrt{GH^2+EH^2}=a-3+\frac{9}{a} EG = G H 2 + E H 2 = a − 3 + a 9 至此，△ A E G \bigtriangleup AEG △ A EG a a a
∴ C △ A E G = 9 \therefore C_{\bigtriangleup AEG}=9 ∴ C △ A EG = 9
(3)
有了前面两问，第三问不是易如反掌？
首先，三角形的内切圆圆心是它的三个角平分线的交点，它到三边的距离相等。其实不知道也可以
所以我们直接拿内切圆半径公式求就行了，r = 2 × S C r=\frac{2\times S}{C} r = C 2 × S
第二问中，我们求出了 C △ E G H C_{\bigtriangleup EGH} C △ EG H A G AG A G E H EH E H a a a a a a a a a S 1 = 3 S 2 S_{1}=3 S_{2} S 1 = 3 S 2 D F G F = 3 1 \frac{DF}{GF}=\frac{3}{1} GF D F = 1 3 D F = 3 4 D G DF=\frac{3}{4}DG D F = 4 3 D G △ D E F \bigtriangleup DEF △ D EF D F = D E DF=DE D F = D E D F = 3 4 D G DF=\frac{3}{4}DG D F = 4 3 D G a a a a a a
最后答案为 5 3 12 \frac{5\sqrt{3}}{12} 12 5 3
