Mixtilinear incircles!!!

WizardMath wrote: Mixtilinear incircles!!! You don't need such advanced things, since $ \angle AID= \angle ABD=90$ we have $AIBD $ cyclic, so $\angle IAB= \angle IDB= \frac{\alpha}{2} $ and $ \angle ABI= \angle ADI=45 $ now we let $ T=CI\cap AD $ and we have $\angle DTC= 180-\frac{\alpha+\gamma}{2}-45=180-45-45=90$ so $CI \perp AD$. By sine law $ AD=\frac{b\cdot \sin(\gamma)}{\cos(\frac{\gamma}{2})}=2b\cdot \sin(\frac{\gamma}{2})=2b\sqrt{\frac{1-\cos (\gamma)}{2}}=b\sqrt{2(1-a/b)}=\sqrt{2b(b-a)}$ since $\frac{AD}{ \sqrt{2}}=DI $ we have the result

can anyone draw the diagram.please.

GoJensenOrGoHome wrote: WizardMath wrote: Mixtilinear incircles!!! You don't need such advanced things, since $ \angle AID= \angle ABD=90$ we have $AIBD $ cyclic, so $\angle IAB= \angle IDB= \frac{\alpha}{2} $ and $ \angle ABI= \angle ADI=45 $ now we let $ T=CI\cap AD $ and we have $\angle DTC= 180-\frac{\alpha+\gamma}{2}-45=180-45-45=90$ so $CI \perp AD$. By sine law $ AD=\frac{b\cdot \sin(\gamma)}{\cos(\frac{\gamma}{2})}=2b\cdot \sin(\frac{\gamma}{2})=2b\sqrt{\frac{1-\cos (\gamma)}{2}}=b\sqrt{2(1-a/b)}=\sqrt{2b(b-a)}$ since $\frac{AD}{ \sqrt{2}}=DI $ we have the result Nice solution. There is a small typo in your first line. It should be \(\angle IAB = \angle IDC\).

GoJensenOrGoHome wrote: WizardMath wrote: Mixtilinear incircles!!! You don't need such advanced things, since $ \angle AID= \angle ABD=90$ we have $AIBD $ cyclic, so $\angle IAB= \angle IDB= \frac{\alpha}{2} $ and $ \angle ABI= \angle ADI=45 $ now we let $ T=CI\cap AD $ and we have $\angle DTC= 180-\frac{\alpha+\gamma}{2}-45=180-45-45=90$ so $CI \perp AD$. By sine law $ AD=\frac{b\cdot \sin(\gamma)}{\cos(\frac{\gamma}{2})}=2b\cdot \sin(\frac{\gamma}{2})=2b\sqrt{\frac{1-\cos (\gamma)}{2}}=b\sqrt{2(1-a/b)}=\sqrt{2b(b-a)}$ since $\frac{AD}{ \sqrt{2}}=DI $ we have the result you actually dont need the law of sines.a simple manipulation of pythagoras' would do. anyways,a nice solution:)

A complete synthetic non-trig solution, just for completeness For the first part, observe that $\angle DBA=\angle DIA(=90^{\circ})$, so $DBIA$ is cyclic, and hence, $\angle IAB=\angle IDB$ and $\angle ADI=\angle BI$. So, $\angle ADB=45+\frac{A}{2}$. Now in $\triangle DEC$, $\angle DEC+\angle ECD+\angle CDE=180^{\circ}\implies \angle DEC=180^{\circ}-45^{\circ}-\frac{A}{2}-\frac{C}{2}=90^{\circ}$, as desired. For the second part, $AI=ID$ since $\angle ADI=45^{\circ}$. Also, $CD=AC$ since $\triangle CED$ and $CEA$ are congruent by ASA criterion. So, $BD=CD-CB=CD-a=b-a$. Observe that $\triangle DBI$ and $\triangle AIC$ are similar, as $\angle BDI=\angle AIC=\frac{A}{2}$ and $\angle DIB=\angle CIA=\frac{C}{2}$. So, $\frac{AI}{BD}=\frac{AC}{ID}\implies\frac{ID}{b-a}=\frac{b}{ID}\implies ID=\sqrt{b(b-a)}$, as desired $\blacksquare$.

Just use cosine rule and isosceles triangles for the 2nd part

Probably the most simple solution to this question with 0 trigonometry

Diagram from Math-wiz

Claim 1. $(ADBI)$ is cyclic. $$\angle AID=\angle AID=\angle ABD$$Claim 2. $AI=ID$ \begin{align*}&45^{\circ}=\angle ABI=\angle ADI\\ &\angle DAI=180^{\circ}-\angle DBA-\angle IBA=45^{\circ}\end{align*}Claim 3. $\triangle CAD$ is isosceles $\implies CI\perp AD$ $$\angle CAD=\angle CAI+\angle IAD=\angle IAB+45^{\circ}=\angle IDB+\angle ADI=\angle ADC$$Now we take $c=AB$ and we have the following $$ID=\sqrt{\frac{AD^2}{2}}=\sqrt{\frac{AB^2-DB^2}{2}}=\sqrt{\frac{c^2-(b-a)^2}{2}}=\sqrt{\frac{2b^2-2ba}{2}}=\sqrt{b(b-a)}$$