I had come across a question in which we had to show that a given quadrilateral, if subjected under a condition, had an incircle. So, will it be sufficient to show that $a+b=c+d$, if $a,b,c,d$ are the sides of the quadrilateral? If yes, then would someone please tell me how the equality of the sum of opposite sides in a quadrilateral necessarily implies the existence of an inscribed circle?
Does the equality of the sum of opposite sides in a quadrilateral necessarily imply the existence of an inscribed circle
circlesgeometryquadrilateral
Related Solutions
This is a complete proof, and it might get a bit repetitive because I'm using the Sine Law again and again. I'm trying to find something more elegant, but this is the only thing thing I could come up with right now.( I've used a different diagram because my internet wasn't working while I was working this out. Just skip over to the ending note if you just want to see how the angles are equal.)
$DEGF$ is the quadrilateral and $A$ and $B$ are the foci.
Using sine law in $\Delta DBE$ you get $\dfrac{DE}{BE}=\dfrac{\sin\angle DBE}{\sin\angle BDE}$ and in $\Delta GBE$ you get $\dfrac{GE}{BE}=\dfrac{\sin\angle GBE}{\sin\angle BGE}$. Dividing the two you'll obtain: $\dfrac{DE}{GE}=\dfrac{\sin\angle DBE}{\sin\angle BDE}\cdot\dfrac{\sin\angle BGE}{\sin\angle GBE}\tag{i}$
Using sine law in $\Delta BDF$ you get $\dfrac{FD}{FB}=\dfrac{\sin\angle DBF}{\sin\angle BDF}$ and in $\Delta BGF$ you get $\dfrac{FG}{FB}=\dfrac{\sin\angle GBF}{\sin\angle BGF}$. Dividing the two you'll obtain: $\dfrac{FD}{FG}=\dfrac{\sin\angle DBF}{\sin\angle BDF}\cdot\dfrac{\sin\angle BGF}{\sin\angle GBF}\tag{ii}$
Dividing (i) and (ii):$\dfrac{DE\cdot FG}{GE\cdot FD}=\dfrac{\sin\angle BGE}{\sin\angle BDE}\cdot\dfrac{\sin\angle BDF}{\sin\angle BGF}\tag{iii}$
Using sine law in $\Delta DFI$ you get $\dfrac{DI}{FI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}$ and in $\Delta GFI$ you get $\dfrac{GI}{FI}=\dfrac{\sin\angle GFE}{\sin\angle FGD}$. Dividing the two you'll obtain: $\dfrac{DI}{GI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}\cdot\dfrac{\sin\angle FGD}{\sin\angle GFE}\tag{iv}$
Using sine law in $\Delta DEI$ you get $\dfrac{DI}{EI}=\dfrac{\sin\angle DEF}{\sin\angle EDG}$ and in $\Delta GEI$ you get $\dfrac{GI}{EI}=\dfrac{\sin\angle GEF}{\sin\angle EGD}$. Dividing the two you'll obtain: $\dfrac{DI}{GI}=\dfrac{\sin\angle DEF}{\sin\angle EDG}\cdot\dfrac{\sin\angle EGD}{\sin\angle GEF}\tag{v}$
Combining (iv) and (v): $\dfrac{DI}{GI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}\cdot\dfrac{\sin\angle FGD}{\sin\angle GFE}=\dfrac{\sin\angle DEF}{\sin\angle EDG}\cdot\dfrac{\sin\angle EGD}{\sin\angle GEF}\tag{vi}$
Doing the same thing with $DB$ and $GB$ you will get $$\dfrac{DB}{GB}=\dfrac{\sin\angle DFE}{\sin\angle GFE}\dfrac{\sin\angle BGF}{\sin\angle BDF}=\dfrac{\sin\angle DEF}{\sin\angle GEF}\dfrac{\sin\angle BGE}{\sin\angle BDE}\tag{vii}$$
Dividing (vi) and (vii) you get:$$\dfrac{\sin\angle FGD}{\sin\angle FDG}\cdot\dfrac{\sin\angle BDF}{\sin\angle BGF}=\dfrac{\sin\angle DGE}{\sin\angle GDE}\cdot\dfrac{\sin\angle BDE}{\sin\angle BGE}\tag{viii}$$
Now due to the Property 1 on this link, $\angle FGD=\angle BGE, \angle GDE=\angle BDF,\angle DGE=\angle BGF$ and $\angle BDE=\angle FDG$. Making the necessary substitutions, relation (viii) becomes:$$\sin^2\angle BGE \cdot\sin^2\angle BDF=\sin^2\angle BDE \cdot\sin^2\angle BGF$$ $$\sin\angle BGE \cdot\sin\angle BDF=\sin\angle BDE \cdot\sin\angle BGF\tag{ix}$$
Combining (iii) and (ix): $DE\cdot FG=GE\cdot FD$
NOTE: Making use of the relations (vi),(vii) and (ix) it is easy to see that $\dfrac{DI}{GI}=\dfrac{DB}{GB}$. By the Angle Bisector Theorem, this means that $\angle DBF = \angle GBF$ and similarly $\angle FAG = \angle EAG$.
OP mentions this blog post as the inspiration for the question. That post might be saying that $AIEF'$ is cyclic because of "radical axis", but if so it makes the error of assuming that $I,D,F'$ are collinear.
The blog post refers to a thread on the problem, of which this is one entry. The entry defines $F'$ as collinear with $I,D$, and in a later step (after showing $AIEF'$ cyclic) demonstrates that $F'$ is collinear with $E,M'$.
Assuming $I,D,F'$ collinear the "radical axis" method amounts to computing powers of the point $D$ with respect to the circles $(ABC)$ and $(BIC)$ and the three lines $AE,BC,IF'$, giving us $$AD\cdot DE=BD\cdot DC=ID\cdot DF'.$$ By the intersecting chords theorem $AD\cdot DE=ID\cdot DF'$ implies $AIEF'$ is cyclic.
After this, we can angle chase (as in the thread entry) to show that $F'$ is collinear with $E,M'$ giving us the harmonic quad setup.
Best Answer
Yes, it's true.
Let $ABCD$ be an equilateral with $AB+CD=BC+AD$.
Also, let $AE$ and $BE$ be bisectors of $\angle BAD$ and $\angle ABC$ respectively.
Thus, the distances from $E$ to $AB$, $AD$ and $BC$ they are equal, which says that the circle with a center $E$ and with a radius is this distance, is touched to sides $AD$, $AB$ and $BC$.
Now, let $CD_1$ is a tangent to the circle, where $D_1$ is placed on the line $AD$.
Thus, $$AB+CD_1=BC+AD_1$$ and $$AB+CD=BC+AD$$ from the given.
Hence, $$CD_1-CD=AD_1-AD$$ or $$CD_1-CD=\pm DD_1,$$ which by the triangle inequality is possible, when $D\equiv D_1$ and we are done!