▻ O'tkir burchakli, teng yonli bo'lmagan $\triangle ABC$ berilgan va uning $AD$, $BE$, $CF$ balandliklari $H$ nuqtada kesishadi. Aytaylik $EF\,\cap \, AD = P(.)$, $BP\, \cap \, (ABC) = B_1(.)$, $CP\, \cap \, (ABC) = C_1(.)$, bu yerda $(ABC)-$ $\triangle ABC$ ning tashqi aylanasi. Agar $B_1C_1 \, \cap \, BC = Q(.)$ va $EF\, \cap \, BC = K(.)$ bo'lsa, $AK$ to'g'ri chiziq $PQ$ kesmani teng ikkiga bo'lishini isbotlang.
-------------------------------------------------------------------
Yechish. Nisbatan, analitik yechim bersakda, boshlanishiga proyektiv boshlaymiz. Quyidagi ikkita qadam masalani to'liq yechimiga olib boradi. 1-qadam. $\triangle PDQ$ ga Menelay teoremasini qo'llash orqali ($\overline{AK}$ - t/ch) shuni topamiz-ki (1-rasm ga qarang): $AK$ to'g'ri chiziq $PQ$ kesmani teng ikkiga ajratadi agar va faqat, agar \[ \frac{QK}{KD} = \frac{AP}{AD} \quad \quad \quad (1) \]nisbat bajarilsa. Lekin boshqa tomondan, bilamiz-ki $\{ A,P,H,D \}$ lar o'zaro harmonik nuqtalar ($AEHF$ to'rtburchakdan tezda bilib olish mumkin) va $\displaystyle \frac{AP}{AD} = \frac{HP}{HD}$, u holda $(1)$ dan $\displaystyle \frac{QK}{KD} = \frac{HP}{HD}$ keladi. Ya'ni, $KH\parallel QP$. Shuningdek, $(BFEC)$ aylana $BC$ diametrli aylanaligidan va $KH$ to'g'ri chiziq $A$ nuqtaning polyar to'g'ri chiziqiligidan, $KH\perp AM$ bo'ladi, bu yerda $M(.)-$ $BC$ kesmaning o'rtasi. Bu degani, oxir oqibatda, $AK$ to'g'ri chiziq $PQ$ kesmani teng ikkiga bo'lishi o'z navbatida $PQ$ ning $AM$ ga perpendikulyarligini ko'rsatish bilan teng kuchliligini bildiradi.
Tasdiq. $QM$ kesma \[ \frac{QM}{BM} = \frac{CP\cdot CC_1 + BP\cdot BB_1}{|CP\cdot CC_1 - BP\cdot BB_1|} \] nisbatni qanoatlantiradi.
Isboti. $\triangle QC_1C $ va $\triangle QB_1B$ lardan
\[\frac{QC}{CC_1} = \frac{\sin \angle QC_1C}{\sin \angle C_1QC}, \quad \quad \quad \frac{BB_1}{QB} = \frac{\sin \angle B_1QB}{\sin \angle QB_1B} \] va \[ \frac{QC}{QB} = \frac{\sin \angle QC_1C}{\sin \angle QB_1B} \cdot \frac{CC_1}{BB_1} = \frac{\sin \angle PBC}{\sin \angle PCB} \cdot \frac{CC_1}{BB_1} = \frac{CP}{BP} \cdot \frac{CC_1}{BB_1} \] ni topamiz. Albatta, bizning holda (1-rasm ga qarang) \[ QC = QM+MC, \quad \quad QB = QM - BM \] va bulardan isbotlanishi kerak bo'lgan nisbatni keltirib chiqarish mumkin. Tasdiq isbotlandi.
-----------------------------------------------------
Yuqoridagi tasdiqdan, shunisi ma'lum-ki ($\mathrm{Pow}(P)$ deganda $P$ nuqtani $(ABC)$ aylanaga nisbatan powerini nazarda tutayapmiz), \[ QM^2 - QD^2 + DM^2 = 2QM \cdot DM = \frac{CP\cdot CC_1 + BP\cdot BB_1}{CP\cdot CC_1 - BP\cdot BB_1} \cdot DM(CD+BD) \] \[ =\frac{CP^2 + BP^2 + 2\mathrm{Pow}(P)}{{CP^2 - BP^2}} \cdot \frac{{(CD^2 - BD^2)}}{2} = \frac{CD^2 + BD^2}{2} + PD^2 + \mathrm{Pow}(P) \] \[ =BM^2 + DM^2 + DP^2 + \mathrm{Pow}(P), \] ya'ni \[ QM^2 - QD^2 = BM^2 + DP^2 + \mathrm{Pow}(P). \quad \quad (2) \]
Aytaylik, $AD \, \cap \, (ABC) = A'$. U holda \[ \mathrm{Pow} (P) = AP\cdot PA' = AP\cdot AA' - AP^2 = AF\cdot AB - AP^2 \] \[ =\mathrm{Pow}_{(BC)}(A) - AP^2 = AM^2 - BM^2 - AP^2 \] keladi, bu yerda $\mathrm{Pow}_{(BC)}(A)$ deganda $A$ nuqtaning $(BC)$ diametrli aylanadagi poweri nazarda tutilmoqda va o'tadagi bitta tenglik $FPA'B$ ning cyclic bo`lgani uchun o'rinli. Ushbu tenglikdan va $(2)$ dan \[ QM^2 - QD^2 = AM^2 + DP^2 - AP^2 \quad \Rightarrow \quad QM^2 - AM^2 = QP^2 - AP^2 \] tenglikni keltirib chiqaramiz. Ko'rishimiz mumkin-ki, oxirgi tenglik $MP\perp AQ$ ga eqvivalentdir. Unda $P$ nuqta $\triangle AMQ$ ning ortomarkazi bo'ladi (chunki, uning balandliklari yotgan to`g`ri chiziqlari $\Rightarrow $ $AD$ va $MP$ lar keltirilgan). Demak, $QP\perp AM$ ekan. Isbot tugadi. ▢