In a right-angled triangle ABC, the height CH drawn from the top of a right angle divided the triangle into two right-angled triangles, each of which contains a circle. Prove that the lines containing the radii of these circles perpendicular to the corresponding hypotenuses (legs of the original triangle) intersect on the hypotenuse of triangle ABC.

This all means that we have a right angle triangle ABC, the height of this triangle is drawn, and in two smaller right triangles are inscribed circles, we need to prove that the perpendicular lines from the centers to the matching legs of the big triangle intersect on the big hypotenuse

I don't know how to prove this, I tried a lot.

  • $\begingroup$ Add a picture... $\endgroup$ – Moti Jul 17 at 5:47

enter image description here

Let $|AB|=c$, $|BC|=a$, $|AC|=b$, $|CD|=d$, $T_1,T_2$ be the touching points, and $X=I_1T_1\cap T_2I_2$.

Then $CT_1XT_2$ is a rectangle and we have \begin{align} d&=\frac{ab}c ,\quad |AD|=\frac{b^2}c ,\quad |BD|=\frac{a^2}c ,\\ |CT_1|=|XT_2|&= \tfrac12\,(b+d-|AD|)= \frac b{2c}(c+a-b) ,\\ |CT_2|=|XT_1|&=\tfrac12\,(a+d-|BD|)= \frac a{2c}(c+b-a) ,\\ |AT_1|&=\tfrac12\,(b+|AD|-d)= \frac b{2c}(c-a+b) ,\\ |BT_2|&=\tfrac12\,(a+|BD|-d)= \frac a{2c}(c+a-b) ,\\ \triangle AXT_1:\quad |AX|&=\sqrt{|AT_1|^2+|XT_1|^2} = \tfrac12\sqrt{\frac{(b+c-a)^2(a^2+b^2)}{c^2}} \\ &= \tfrac12(b+c-a) ,\\ \triangle BXT_2:\quad |BX|&=\sqrt{|BT_2|^2+|XT_2|^2} = \tfrac12\sqrt{\frac{(a+c-b)^2(a^2+b^2)}{c^2}} \\ &=\tfrac12(a+c-b) ,\\ |AX|+|BX|&=c=|AB| , \end{align}

hence $\triangle ABX$ is degenerate and $X\in AB$.

Note that the point $X$ is in fact a touching point of the incircle of $\triangle ABC$.

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  • $\begingroup$ How do you use the similarity of triangles if you do not know that $X$ is on $AB$ . Maybe $X$ is not there at all. One could assume that $X_1$ and $X_2$ are the intersection points of $T_1 I_1$ of line AB and $T_2 I_2$ of line AB, respectively then you will show that $BX_2+AX_1=c$ so $X_2$ and $X_1$ are the same point $\endgroup$ – Delta Account Jul 17 at 9:53
  • $\begingroup$ @Delta Account: thanks, you are right, I'll try to fix this asap. $\endgroup$ – g.kov Jul 17 at 10:06
  • $\begingroup$ @Delta Account: updated. $\endgroup$ – g.kov Jul 17 at 10:42


So, we draw the incircle in one $\Delta AHC$ and let its incircle touch its hypotenuse at point $G$, then let the line $GE \perp AC$ through its incenter $E$ intersect the hypotenuse of $\Delta ABC$ at $\alpha$.
$\alpha K \perp BC$ is drawn, intersecting $BC$ at $K$. We will prove that the incircle of $\Delta BHC$ touches $BC$ precisely at $K$, which is same as proving the required statement.

Proof: We note that $\Delta AHC \sim \Delta ACB, \ \because$ they share $\angle A$ and a right angle. ($\Delta BHC \sim \Delta BCA$ since they share $\angle B$ and a right angle)
We have, by similarity in the first pair of triangles, $\frac{AH}{AC}=\frac{AC}{AB} \implies AH = \frac{b^2}c$
(Using $AB=c,BC=a,CA=b$)
and by similarity in the second pair of triangles, $BH=\frac{a^2}{c}$.
We can find the length of $CH$ by considering $$\text{area}[ABC]=\frac12 CH\cdot c \implies \frac12ab=\frac12 CH\cdot c \implies CH = \dfrac{ab}c$$

Thus we know the lengths of all three sides of $\Delta ACH$. The length $AG$ can be worked out from the formula known for the distance between the vertex and the nearest touchpoint of the incircle. as $$AG = \frac{AC+AH-CH}2=\frac{b}c(s-a) \ \text{where } s=\frac{a+b+c}2$$

I have done the work of showing how to derive required lengths. Following the outline below, you are encouraged to complete the rest of the proof.

$(1)$ $G\alpha || CB \implies \Delta AG\alpha \sim \Delta ACB$, and using this find the length of $G\alpha$ in terms of $a,b,c$.
$(2)$ Observe that $G\alpha KC$ is a rectangle, so $G\alpha = CK$.
$(3)$ In a fashion exactly similar to finding the length of $BG$, assume that the incircle of $\Delta BHC$ touches $BC$ at $K'$ and find the length of $CK'$.
$(4)$ Conclude from $(2),(3)$ that $CK=CK' \implies K=K'$.

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