is there a mistake in b

Well; for 1 Notice that $\angle EFG = 180 - \angle EFB - \angle GFC$. But quadrilaterals $EOFB$ and $OFCG$ are cyclic (by the perpendiculars) so $\angle EFB = 90 - \angle OBE$ and $\angle GFC = 90 - \angle OCG$ and thus $\angle EFG = \angle OBE + \angle GFC$. Doing the same for angle $\angle GHE$ will get that the required sum is $\angle OBE + \angle OCG + \angle ODC + \angle OAE = 180$ Edit: Looked this up on google; see theorem 3 http://forumgeom.fau.edu/FG2012volume12/FG201202.pdf

ythomashu wrote: is there a mistake in b Indeed. The problem amounts to showing that $ABCD$ is cyclic, which is not always true for orthodiagonal quadrilaterals.

the original formulation was Quote: Diberikan segiempat talibusur $ABCD$ dengan kedua diagonalnya saling tegak lurus dan berpotongan di titik $O$. Garis tegak lurus dari $O$ pada $AB$; memotong $AB$ di $E$. Garis tegak lurus dari $O$ pada $BC$, memotong $BC$ di $F$. Garis tegak lurus dari $O$ pada $CD$, memotong $CD$ di $G$: Garis tegak lurus $O$ pada $DA$, memotong $DA$ di $H$. a. Buktikan bahwa $\angle EFG + \angle GHE = 180^o$ b. Buktikan bahwa $OE$ merupakan garis bagi sudut $FEH$. I used google translate and changed the wording in the start, in order to make it more simple PS. official solution is available

parmenides51 wrote: the original formulation was Quote: Diberikan segiempat talibusur $ABCD$ dengan kedua diagonalnya saling tegak lurus dan berpotongan di titik $O$. Garis tegak lurus dari $O$ pada $AB$; memotong $AB$ di $E$. Garis tegak lurus dari $O$ pada $BC$, memotong $BC$ di $F$. Garis tegak lurus dari $O$ pada $CD$, memotong $CD$ di $G$: Garis tegak lurus $O$ pada $DA$, memotong $DA$ di $H$. a. Buktikan bahwa $\angle EFG + \angle GHE = 180^o$ b. Buktikan bahwa $OE$ merupakan garis bagi sudut $FEH$. I used google translate and changed the wording in the start, in order to make it more simple PS. official solution is available can you please send it? maybe we can decipher backwards and understand the problem?

the official solution

Attachments:
2016 inamo p1 solved.pdf (85kb)

On what basis are they assuming that $\angle BCD =\angle CAD$ which basically means the quadrilateral is cyclic? Someone who understands Bahasa/Malay can help please?

achen29 wrote: On what basis are they assuming that $\angle BCD =\angle CAD$ which basically means the quadrilateral is cyclic? Someone who understands Bahasa/Malay can help please? Hello, this should help: "Diberikan segiempat talibusur $ABCD$..." is translated to : "Given cyclic quadrilateral $ABCD$...$

Ronald Widjojo wrote: Hello, this should help: "Diberikan segiempat talibusur $ABCD$..." is translated to : "Given cyclic quadrilateral $ABCD$... thank you, I am gonna correct it, my original post was parmenides51 wrote: Let $ABCD$ be a quadrilateral wih both diagonals perpendicular to each other and intersecting at point $O$. Let $E,F,G,H$ be the orthogonal projections of $O$ on sides $AB,BC,CD,DA$ respectively. a. Prove that $\angle EFG + \angle GHE = 180^o$ b. Prove that $OE$ bisects angle $\angle FEH$ . As it seems I missed the cyclic condition at the start, and the second question was not valid, sorry

First, note that $OEBF, OEAH, OHDG$ and $OFGC$ are cyclic. Hence, \begin{align*} \angle EHG+\angle EFG\\&=\angle EHO+\angle GHO+\angle EFO+\angle GFO\\&=\angle EAO+\angle GDO+\angle EBO+\angle GCO\\&=\angle EAC+\angle GDB+\angle EBD+\angle GCA\\&=2(\angle CAB+\angle ABD)\\&=2(180^{\circ}-90^{\circ})\\&=180^{\circ},\end{align*}proving part a. For part b, note that $$ \angle HEO=\angle HAO=\angle DAC=\angle DBC=\angle OBF=\angle OEF,$$proving that $OE$ bisects $\angle FEH.$ $\blacksquare$