*Sharygin Geometry Olympiad, Russia*

## Inscribed-Quad in an Excribed-Quad

### Inscribed-Quad in an Excribed-Quad

The quadrilateral $ABCD$ is excribed around a circle with centre $I$. Prove that, the projections of $B$ and $D$ on $IA$, $IC$ lie on a circle.

Last edited by sowmitra on Mon May 04, 2015 8:38 pm, edited 1 time in total.

### Re: Inscribed-Quad in an Excribed-Quad

You sure about the statement?

**- What is the value of the contour integral around Western Europe?**

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

### Re: Inscribed-Quad in an Excribed-Quad

Extremely sorry for the typo...

### Re: Inscribed-Quad in an Excribed-Quad

Points $W,X$ are projections of $B$ and $Y,Z$ are projections of $D$ on $IA$ and $IC$ respectively. Multiple configs can occur, so angles are directed mod $\pi$.

Since $\measuredangle IWB=\measuredangle IXB=\pi/2$ quad $IWBX$ is cyclic. Similarly quad $IZDY$ is cyclic.

So we have \[\measuredangle WXI = \measuredangle WBI=\pi/2 - \measuredangle BIW=\pi/2+\measuredangle ABI+\measuredangle IAB=\frac{1}{2}\left(\pi+\measuredangle ABC+\measuredangle DAB\right)\] and similarly $\measuredangle IYZ=\frac{1}{2}\left(\pi+\measuredangle DCB+\measuredangle ADC\right)$.

Hence $WXYZ$ cyclic iff \[\measuredangle WXI=\measuredangle IYZ\Leftrightarrow \measuredangle ABC+\measuredangle DAB=\measuredangle DCB+\measuredangle ADC\] \[\Leftrightarrow \measuredangle ADC+\measuredangle DCB+\measuredangle CBA+\measuredangle BAD=0\] which is true for quad $ABCD$.

**- What is the value of the contour integral around Western Europe?**

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

### Re: Inscribed-Quad in an Excribed-Quad

i could not visit the forum last 2 days for internet troubles . so didn't notice the typo update .

was trying to solve the previous .....then hssss

whatever, now i got the solution which nirjhor posted

was trying to solve the previous .....then hssss

whatever, now i got the solution which nirjhor posted

### Re: Inscribed-Quad in an Excribed-Quad

I have got another solution.I am writing it shortly:

Suppose,$BX$ and $BW$ are perpendiculars to $CI$ and $AI$ respectively and suppose,$DZ$ and $DY$ are perpendiculars to $CI$ and $AI$.

$\measuredangle AID+\measuredangle BIC=0^{0}$.

$\measuredangle BIC+\measuredangle BIX=0^{0}$.

$\therefore \measuredangle BIX=\measuredangle AID$.

$\therefore \measuredangle XBI=\measuredangle XWI=\measuredangle YDI=\measuredangle YZI$.

$\therefore WXYZ$ is cyclic.

Suppose,$BX$ and $BW$ are perpendiculars to $CI$ and $AI$ respectively and suppose,$DZ$ and $DY$ are perpendiculars to $CI$ and $AI$.

$\measuredangle AID+\measuredangle BIC=0^{0}$.

$\measuredangle BIC+\measuredangle BIX=0^{0}$.

$\therefore \measuredangle BIX=\measuredangle AID$.

$\therefore \measuredangle XBI=\measuredangle XWI=\measuredangle YDI=\measuredangle YZI$.

$\therefore WXYZ$ is cyclic.

"Questions we can't answer are far better than answers we can't question"