Prove that D is incenter of ABC

Question

Question - AB,AC are tangents from A to a circle touching it at B and C. if D is midpoint of minor arc BC prove that D is incentre of ABC.

My try - First I proved that A,D,O are collinear using given condition that D is midpoint of arc BC and tangents from external point subtends equal angle at centre..

So AD bisects angle A ... Now I drop perpendicular DE,DF,DG from D to sides AB,BC,CA respectively..

Now I am able to prove that DE=DG by congruency but after applying all ideas that I have I am not able to show that DE=DG=DF .....

Any help will be greatly helpful using Euclidean geometry

Thankyou

Answer 1

You already proved that $DE=DG$, so we want to prove that $DE=DF$.

$\quad$

First, since $\angle{OBA}=\angle{OFB}\ (=90^\circ)$, we see that $\triangle{OBA}$ and $\triangle{OFB}$ are similar, so we get $$\angle{OAB}=\angle{OBF}\tag1$$ Since $\triangle{OBD}$ is an isosceles triangle with $OB=OD$, we get $$\angle{ODB}=\angle{OBF}+\angle{FBD}\tag2$$ Also, we get $$\angle{ODB}=\angle{OAB}+\angle{DBA}\tag3$$

It follows from $(1)(2)(3)$ that $$\angle{FBD}=\angle{DBA}\tag4$$

From $(4)$, we see that $\triangle{DBF}$ is congruent to $\triangle{DBE}$, from which $$DE=DF$$ follows.

Answer 2

A certain case: As can be see in drawing, If $AD=R$, where $R $ is the radius of circle , then D will be the in-center of triangle.

Answer 3

Let $|IB_t|=|IC_t|=r$, $|AB_t|=|AC_t|=t$, $|DE|=|DG|=r_t$.

Then

\begin{align} \angle A_tAB&=\arctan\tfrac rt=\tfrac\alpha2 ,\\ \sin\tfrac\alpha2&=\frac r{\sqrt{t^2+r^2}} ,\quad \cos\tfrac\alpha2=\frac t{\sqrt{t^2+r^2}} . \end{align}

\begin{align} \frac{r_t}r &= \frac{|AH|+2\,r_t}{|AH|+r_t+2\,r} =\frac{|AH|}{|AH|+r_t} ,\\ r_t&= \tfrac12\,|AH|\cdot\left(-1+\sqrt{1+\frac{4\,r}{|AH|}}\right) ,\\ |AH|&= \frac{r_t}{\sin\tfrac\alpha2}-r_t = \frac{r_t\,\sqrt{r^2+t^2}}r-r_t . \end{align}

\begin{align} r_t&= r\,\left(1-\frac r{\sqrt{t^2+r^2}}\right) ,\\ \frac{r_t}r&= 1-\frac{r}{\sqrt{t^2+r^2}} . \end{align}

\begin{align} \triangle AC_tF:\quad |AF|&=t\,\cos\tfrac\alpha2 = \frac{t^2}{\sqrt{t^2+r^2}} \tag{1}\label{1} . \end{align}

On the other hand,

\begin{align} |AH|+2\,r_t&= r_t+\frac{r_t}r\,\sqrt{r^2+t^2} \\ &=r_t-r+\sqrt{t^2+r^2} \\ &= r\,\left(1-\frac r{\sqrt{t^2+r^2}}\right)-r+\sqrt{t^2+r^2} \\ &=\frac{t^2}{\sqrt{t^2+r^2}} \tag{2}\label{2} . \end{align}

Since \eqref{1}$=$\eqref{2}, $|AF|=|AH|+2\,r_t$ and $D$ is indeed the incenter of $\triangle AC_tB_t$.

Answer 4

Since $D$ is the midpoint of the arc $BC$, $\stackrel \frown {BD}\> = \>\stackrel \frown {DC}$. Then, their subtended angles on the circle $\angle EBD = \angle FBD$. Alone with the shared side $BD$, the right triangles $\triangle EBD$ and $\triangle FBD$ are congruent.

Thus, $FD = DE$. Similarly, $FD = DG$, hence $D$ being the incenter.

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For completeness, the proof of the collinearity of $A$, $D$ and $O$, as well as $BC \perp AF$, is provided here, even though you indicated you had already done so. Given the tangents $AB$ and $AC$, the right-angle $\triangle ABO$ and $\triangle ACO$ are congruent due to $OB = OC$ and the shared $OA$. Then, $\angle BAF= \angle CAF$, $AB = AC$. With the shared $AF$, the $\triangle ABF$ and $\triangle ACF$ are congruent. Then, $BC \perp AF$. Also, from $\angle BOA= \angle COA$, $AO$ intersects the circle at $D$.

Answer 5

I'm sorry I can't get the pictures for the explanations right now but in the meantime I can leave as an exercise to the viewer to prove the similarity of the triangles I'll be using.

My main weapons were concepts of similarity and the plain fact that $|OA|=|OB|=r$

So by similarity of $\triangle OFA$ and $\triangle OAC$:

$$\frac{|OF|}{|OA|}=\frac{|OA|}{|OC|}$$

$$|OF| \cdot |OC|=|OA|^2=r^2$$

Again, by similarity of $\triangle DEC$ and $\triangle OFA$:

$$\frac{|DE|}{|DC|}=\frac{|OF|}{|OA|}$$

$$|OA| \cdot |DE|=|OF| \cdot |DC|$$

$$r \cdot |DE|=|OF| \cdot |DC|$$

$$|DC|=\dfrac{r \cdot |DE|}{|OF|}$$

Back to the first equation we formed:

$$|OF| \cdot (|OD|+|DC|)=r^2$$

$$r \cdot |OF| + |OF| \cdot |DC|=r^2$$

$$r \cdot |OF| + |OF| \cdot \dfrac{r \cdot |DE|}{|OF|}=r^2$$

$$r(|OF|+|DE|)=r^2$$

$$|OF|+|DE|=r=|OD|$$

But it has already been shown that $O,F,D$ are collinear thus:

$$|OF|+|DF|=|OD|$$

$$|DF|=|OD|-|OF|=|DE|$$

Therefore $|DE|=|DF|$

Euclid is one of my favorite mathematicians so I hope this answer did him justice.