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For reference: In a semicircle of diameter $AC$, a triangle $ABC$ is inscribed, the points are joined averages of $\overset{\LARGE{\frown}}{AB}$, and $\overset{\LARGE{\frown}}{BC}$ with the vertices $C$ and $A$ that intersect at points $E$ and $F$ with sides $AB$ and $BC$ respectively. Then we draw $EH$ and $FG$ perpendicular to $AC$. Calculate the radius of the circle inscribed in triangle $ABC$ if $HG = 4m$.

My Progress: enter image description here

I made the drawing above(without scale). Relationships I found: $\triangle ABC$ is a rectangle. $\triangle AJH \sim \triangle AMI \sim \triangle AFG$ I think some data is missing...

this is the correct picture enter image description here

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    $\begingroup$ "the points are joined averages of AB, and BC with the vertices C and A that intersect at points E and F with sides AB and BC respectively" is hardly understandable. From the figure I gather that $E$ and $F$ are the midpoints of arcs $AB$ and $BC$, right? $\endgroup$ Commented Sep 1, 2021 at 19:27
  • $\begingroup$ @Intelligentipauca ,,by geogebra is your statement correct..."E and F are the midpoints of arcs AB and BC" $\endgroup$ Commented Sep 1, 2021 at 20:26
  • $\begingroup$ I did the wrong design...I'm correcting $\endgroup$ Commented Sep 18, 2021 at 17:10

2 Answers 2

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If I understand correctly, $E$ is on the segment $\overline{AB}$ and the line $\overleftrightarrow{CE}$ intersects the arc $\overset{\large\frown}{AB}$ at the midpoint of the arc; $F$ is on the segment $\overline{BC}$ and the line $\overleftrightarrow{AF}$ intersects the arc $\overset{\large\frown}{BC}$ at the midpoint of the arc.

Therefore $\overrightarrow{CE}$ is the bisector of the angle $\angle ACB$ and $\overrightarrow{AF}$ is the bisector of the angle $\angle BAC.$ Therefore $\overrightarrow{CE}$ and $\overrightarrow{AF}$ intersect at $D,$ the center of the inscribed circle, as shown in your figure.

Because $\overrightarrow{CE}$ and $\overrightarrow{AF}$ are angle bisectors, because $\angle ABC$ is a right angle, and because $\overline{EH}$ and $\overline{FG}$ are perpendicular to $\overline{AC},$ it follows that $\triangle ABF \cong \triangle AGF$ and $\triangle CBE \cong \triangle CHE.$

The relationships of the segments $\overline{EH}$ and $\overline{FG}$ to the inscribed circle should then be obvious (namely, they are exactly as they appear in your figure). Then the relationship of the distance $GH$ to the radius of the inscribed circle is easily seen.

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  • $\begingroup$ Very good resolution...thanks...So the inraio is equal to $\frac{HG}{2}? $What guarantees that EH and FG are tangent to the circumference? $\endgroup$ Commented Sep 19, 2021 at 11:21
  • $\begingroup$ Consider $\triangle ABF$ and $\triangle AGF.$ Not only are they congruent, one is the mirror image of the other across the line $AF.$ The inscribed circle is its own mirror image across the line $AF,$ so it has the same relationship to $\triangle ABF$ as to $\triangle AGF.$ Alternatively, you could drop perpendiculars from $D$ to the segments $BF$ and $FG$ and show that this produces two more congruent right triangles, each of which has a leg that is a radius of the inscribed circle. $\endgroup$
    – David K
    Commented Sep 19, 2021 at 14:03
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Let $\measuredangle BAC=\alpha.$

Thus, $$AG=2R\cos^2\frac{\alpha}{2}$$ and $$CE=2R\cos^2\left(45^{\circ}-\frac{\alpha}{2}\right),$$ which gives $$4m=HG=AG+CE-AC=2R\left(\cos^2\frac{\alpha}{2}+\cos^2\left(45^{\circ}-\frac{\alpha}{2}\right)-1\right)=$$ $$=R(\cos\alpha+\sin\alpha)=\frac{1}{2}r\left(\cot\frac{\alpha}{2}+\cot\left(45^{\circ}-\frac{\alpha}{2}\right)\right)(\cos\alpha+\sin\alpha)=$$ $$=\frac{r\sin45^{\circ}(\cos\alpha+\sin\alpha)}{2\sin\frac{\alpha}{2}\sin\left(45^{\circ}-\frac{\alpha}{2}\right)}=\frac{r(\cos\alpha+\sin\alpha)}{\sqrt2\left(\cos(45^{\circ}-\alpha)-\cos45^{\circ}\right)}=\frac{r(\cos\alpha+\sin\alpha)}{\cos\alpha+\sin\alpha-1}$$ and we see that something missed in the given.

We got also $$4m=\frac{1}{2}(AB+BC)=\frac{1}{2}(AB+BC-AC)+\frac{1}{2}AC=r+R.$$

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  • $\begingroup$ the picture was not correct..posted it corrected $\endgroup$ Commented Sep 18, 2021 at 17:15

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