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Category: Geometry

Question-16464

Question Number 16464 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17 $${A}\overset{\Delta} {{B}C},{is}\:{equilateral}\:{and}\:'{K}'\:{is}\:{a}\:{point} \\ $$$${on}\:{circumcircle}\:\:{of}\:{it}.{draw}\:{KH},{KJ},{KL} \\ $$$${parallel}\:{to}\:{ABC}'{s}\:{sides}\:{as}\:{showen}. \\ $$$$\left.\mathrm{1}\right){prove}\:{that}: \\…

Question-16409

Question Number 16409 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Jun/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Jun/17 $$\Delta{ABC},{is}\:{equilateral}\:{and}\:'{D}',{is}\:{a}\:{point}\:{on}\:{circumcircle}\:{of}\:{A}\overset{\Delta} {{B}C}. \\ $$$${prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{DA}={DB}+{DC}. \\ $$ Commented…

Question-81887

Question Number 81887 by lalitchand last updated on 16/Feb/20 Commented by mr W last updated on 16/Feb/20 $${let}\:\frac{{FA}}{{BA}}={k},\:\frac{\mathrm{1}}{\mathrm{2}}<{k}<\mathrm{1} \\ $$$${DF}=\frac{{BA}}{\mathrm{2}} \\ $$$$\frac{{OD}}{{OF}}=\frac{{DF}}{{FA}}=\frac{{BA}}{\mathrm{2}×{k}×{BA}}=\frac{\mathrm{1}}{\mathrm{2}{k}} \\ $$$${i}.{e}.\:\frac{{OD}}{{OD}+{DF}}=\frac{\mathrm{1}}{\mathrm{2}{k}} \\…

Related-to-Q16140-What-if-the-three-lines-d-1-d-2-d-3-are-not-parallel-but-concurrent-

Question Number 16302 by mrW1 last updated on 20/Jun/17 $$\mathrm{Related}\:\mathrm{to}\:\mathrm{Q16140} \\ $$$$\mathrm{What}\:\mathrm{if}\:\mathrm{the}\:\mathrm{three}\:\mathrm{lines}\:\mathrm{d}_{\mathrm{1}} ,\mathrm{d}_{\mathrm{2}} ,\mathrm{d}_{\mathrm{3}} \:\mathrm{are} \\ $$$$\mathrm{not}\:\mathrm{parallel},\:\mathrm{but}\:\mathrm{concurrent}? \\ $$ Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on…

Question-16277

Question Number 16277 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17 $${in}\:{triangle}:{A}\overset{\Delta} {{B}C},{consteact}\:\mathrm{3}\:{equilateral} \\ $$$${triangle}\:{on}\:{each}\:{sides}. \\ $$$${prove}\:{that}:\:\:{AL}={BM}={CN}\:. \\ $$ Answered…