In a rhombus $ABCD$, prove that $IG\perp IP$.

Question

Let $ABCD$ be a rhombus with $\angle ADC=60^\circ$ (picture in attach file). The points $E$, $F$, $G$, and $H$ are midpoints on sides $AB$, $DA$, $CD$, and $BC$, respectively. Let $J$ be the intersection of $FH$ and $AG$, and $I$ is midpoint of $FJ$. Prove that $IG \perp IP$.

Answer 1

Let $a$ denote $AB=BC=CD=DA$. Then, $\triangle AOF$ is equilateral with side length $\frac{a}{2}$. That is, $OJ=FJ=\frac{a}{4}$, and $JI=\frac{a}{8}$. Hence, $OI=OJ+JI=\frac{3a}{8}$. Now, $BO=\frac{\sqrt{3}a}{2}$, so $PO=\frac{BO}{2}=\frac{\sqrt3a}{4}$. Since $\angle POI=\angle POA+\angle AOF=90^\circ+60^\circ=150^\circ$, we have form the cosines' law that $$PI^2=PO^2+OI^2-2\ PO\ OI\ \cos(\angle POI)=\frac{3a^2}{16}+\frac{9a^2}{64}-2\left(\frac{\sqrt{3}{a}}{4}\right)\left(\frac{3a}{8}\right)\left(-\frac{\sqrt{3}}{2}\right).$$ So, $PI^2=\frac{39a^2}{64}$.

By Pythagoras, $\triangle IJG$ is a right triangle with $JG=\frac{AG}{2}=\frac{BO}{2}=\frac{\sqrt{3}a}{4}$, so $$IG^2=IJ^2+JG^2=\left(\frac{a}{8}\right)^2+\left(\frac{\sqrt{3}a}{4}\right)^2=\frac{13a^2}{64}.$$ Similarly, $\triangle PHG$ is a right triangle with $PH=\frac{EH}{2}=\frac{AC}{4}=\frac{a}{4}$ and $HG=BO=\frac{\sqrt{3}a}{2}$, so that $$PG^2=\left(\frac{a}{4}\right)^2+\left(\frac{\sqrt{3}a}{2}\right)^2=\frac{13a^2}{16}.$$ Note that $$PI^2+IG^2=\frac{39a^2}{64}+\frac{13a^2}{64}=\frac{13a^2}{16}=PG^2.$$ By the converse to Pythagoras' theorem, $\angle PIG=90^\circ$.

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Alternatively,let $FG$ meet $OD$ at $M$. Since $M$ bisects $FG$ and $I$ bisects $FJ$, $IM$ is parallel to $JG$. That is, $IM$ is perpendicular to $IH$, since $JG\perp OF$. That is $\angle HIM=90^\circ=\angle HPM$. That is $\square HPIM$ is cyclic. This means $$\angle PIH=\angle PMH=\angle PGH.$$ Thus, $\square PIGH$ is also cyclic. This implies $$\angle PIG=180^\circ-\angle PHG=180^\circ-90^\circ=90^\circ.$$

Answer 2

Let $O=(0,0)$, $A = (0,2)$ and $D = (2\sqrt{3},0)$.

Then $P = (-\sqrt{3},0)$, $H= (-\sqrt{3},-1)$, $G= (\sqrt{3},-1)$ and $F = (\sqrt{3},1)$.

Now the line $AG$ is $y= -\sqrt{3}x+2$ and the line $HF$ is $y= {\sqrt{3}\over 3}x$ and they meet at $$J=({\sqrt{3}\over 2},{1\over 2})\implies I = ({3\sqrt{3}\over 4},{3\over 4})$$

Now the slope of $PI$ is $k={\sqrt{3}\over 7}$ and the slope of $IG$ is $k' =-{7\over \sqrt{3}} $ so $k\cdot k'=-1$ and thus $\angle PIG = 90^{\circ}$.

Answer 3

Since $HF||CD$ and $AG\bot CD$ we have that $AG\bot CD$. Now you can use this (observe only rectangular $EFGH$):

Let $ABCD$ be a rectangle, $E$ midpoint of $DC$ and $G\in AC$ such that $BG\bot AC$. Let $F$ be a midpoint of $AG$. Prove $\angle BFE =\pi/2$.

or

Proving right angle using vectors