What is the radius of the circle inscribed in triangle ABC?

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:

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

• "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? Commented Sep 1, 2021 at 19:27
• @Intelligentipauca ,,by geogebra is your statement correct..."E and F are the midpoints of arcs AB and BC" Commented Sep 1, 2021 at 20:26
• I did the wrong design...I'm correcting Commented Sep 18, 2021 at 17:10

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.

• Very good resolution...thanks...So the inraio is equal to $\frac{HG}{2}?$What guarantees that EH and FG are tangent to the circumference? Commented Sep 19, 2021 at 11:21
• 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. Commented Sep 19, 2021 at 14:03

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

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