A unique solution for finding the length y? | (Fun Geometry Problem) |

[@Juan.alende]

Nice

[@MathandEngineering]

Thanks

[@xualain3129]

Let D be the point on AC and BD perpendicular to AC. tan(A)=8/AB=4/x we get AB=2*x From bisector angle theorem of A We have AB/x=AD/z=2=cot(A/2). From which we can get sin(A/2)=1/sqrt(5) and cos(A/2)=2/sqrt(5) y=8*sin(A)=8*2*sin(A/2)*cos(A/2)=32/5=6.4

[@MathandEngineering]

Wow, this really saves tim and is also precise, I didn't see any errors and, there were no round offs, 6.4 is a 100% accurate answer

[@SuchaiSuteparuk]

Let the perpendicular line from B to AC is BDF The point between BC is E. So DE = 4 and FC = y Since AD is the angle bisector of FAB, so AF/AB = FD/BD * but triangle AFB is similar to triangle BFC. So AF/BF = AB/BC --> AF/AB = BF/BC ** From * & ** FD/BD = BF/BC FD/BF = BD/BC 1 - FD/BF = 1 - BD/BC (BF - FD)/BF = (BC - BD)/BC BD/BF = (BC - BD)/BC # But triangle BDE is similar to triangle BFC. So BD/BF = BE/BC ## From # and ## (BC - BD)/BC = BE/BC BC - BD = BE --> 8 - BD = BE In triangle BDE, by Pythagoras BE^2 = BD^2 + DE^2 (8 - BD)^2 = BD^2 + 4^2 64- 16BD + BD^2 = BD^2 + 16 BD = 3 Then BE = 8 - 3 = 5 Triangle BFC is similar to BDE FC/DE = BC/BE y/4 = 8/5 *** Answer y = 6.4 *** We can also find the length of the rest. AB = 6, AF = 3.6, DF = 1.8

[@MathandEngineering]

Wow even though I have not gotten to go through it intensively, but I am really impressed that you didn't have to solve any Quadratic equation, I had also tried to solve it with a method that doesn't involve any Quadratic equation, but I guess I didn't give it enough time, thanks I love the method

[@salmankhan-st6oz]

Thanks you sir, i have really learnt how to solve these kind of math problem alot from your channel

[@MathandEngineering]

Thank you