Geometry

[Guest]

In [tex]\triangle[/tex]ABC; CD,CF, CE divide [tex]\angle[/tex]C into 4 equal parts. CD [tex]\bot[/tex]AB, E is midpoint of AB. Find angle A,B,C . ( Please use congruent triangles; alternate, corresponding angle; equilateral and isosceles triangle; centroid, circumcenter,incenter, orthocenter)

[Guest]

What grade are you in ?

Through trigonometry i got [tex]\angle[/tex]A=([tex]\frac{45}{2}[/tex])[tex]^\circ[/tex] ,[tex]\angle[/tex]B=([tex]\frac{135}{2}[/tex])[tex]^\circ[/tex] ,[tex]\angle[/tex]C=90[tex]^\circ[/tex]

[Guest]

Hi! I'm the author of this post. For your question, I'm currently in grade 7,i got this question as a challenge from my teacher. I've been trying to solve this for 2 weeks but made no progress.Your answer gave me a lot of clues in solving the problem. Thank you very much for your contributions.
( Adds : Side-angle relations, triangle inequality, midlines theorem, exterior angle theorem are allowed).

[Guest]

My decision is long ,I'II send it tomorrow .

[Guest]

Let's start BC=a ,AC=b ,[tex]\angle[/tex]ACE=[tex]\angle[/tex]ECF=[tex]\angle[/tex]FCD=[tex]\angle[/tex]DCB=[tex]\varphi[/tex] .
1. CD=CD
2. [tex]\angle[/tex]FDC=[tex]\angle[/tex]BDC=90[tex]^\circ[/tex]
3. [tex]\angle[/tex]FCD=[tex]\angle[/tex]DCB=[tex]\varphi[/tex] [tex]\Rightarrow[/tex] 2 symptom [tex]\triangle[/tex]FDC[tex]\cong \triangle[/tex]BDC [tex]\Rightarrow[/tex] FC=a
CE is an angle bisector in [tex]\triangle[/tex]AFC [tex]\Rightarrow[/tex] [tex]\frac{AE}{EF}= \frac{AC}{FC}[/tex] ;[tex]\frac{AE}{EF} =\frac{b}{a}[/tex] ;let AE=bp ,EF=ap

CF is an angle bisector in[tex]\triangle[/tex]ABC [tex]\Rightarrow[/tex] two conclusions :
first [tex]\frac{AF}{BF}= \frac{AC}{BC}[/tex] ; [tex]\frac{ap+bp}{bp-ap} =\frac{b}{a} ; \frac{a+b}{b-a}= \frac{b}{a}[/tex]
[tex]b^{2 }-2ab- a^{2 }[/tex]= 0 (1)

[Guest]

second - The formula is valid [tex]CF^{2 }[/tex]=AC.BC-AF.BF
[tex]a^{2 }[/tex]=ba-(ap+bp)(bp-ap) ; [tex]a^{2 }[/tex]=ab-[tex]p^{2 }( b^{2 } -a^{2 }[/tex]) ;[tex]p^{2 }= \frac{ab- a^{2 } }{ b^{2 } -a^{2 } }[/tex] ; [tex]p^{2 }= \frac{a}{a+b}[/tex] (2)

[Guest]

We check if the formula is valid [tex]h_{c } ^{2 }= a_{1 } b_{1 }[/tex] (3)
According to our drawing [tex]CD^{2 }[/tex]=AD.BD

[tex]CF^{2 } -FD^{2 }[/tex]= ? (2bp- -[tex]\frac{p(b-a)}{2} ).[/tex] [tex]\frac{p(b-a)}{2}[/tex]

[Guest]

[tex]a^{2 } - ( \frac{p(b-a)}{2} )^{2 }[/tex] =? 2bp.[tex]\frac{p(b-a)}{2}[/tex] - [tex][ \frac{p(b-a)}{2} ]^{2 }[/tex]

... ... ...
[tex]a^{2 } = ? b^{2 } p^{2 } -ab .p^{2 }[/tex] (see (2) )
[tex]a^{2 } = ? b^{2 } . \frac{a}{a+b} -ab. \frac{a}{a+b}[/tex] ; [tex]a^{2 } = ? \frac{ab(b-a)}{a+b}[/tex] ;[tex]a^{2 } +ab = ? b^{2 } -ab[/tex]
[tex]b^{2 }-2ab- a^{2 }[/tex] = ? 0
This is true according to (1)
If formula (3) is valid ,then [tex]\triangle[/tex]ABC is rectanglular with [tex]\angle[/tex]ACB=90[tex]^\circ[/tex]

[Guest]

[tex]\angle[/tex]ACB=4[tex]\varphi[/tex]= 90[tex]^\circ[/tex] ;[tex]\varphi[/tex]=( [tex]\frac{45}{2}[/tex])[tex]^\circ[/tex]
([tex]\triangle[/tex]DBC -rectangular) [tex]\angle[/tex]B=90[tex]^\circ[/tex]-[tex]\varphi[/tex] =([tex]\frac{180}{2}[/tex])[tex]^\circ[/tex] -([tex]\frac{45}{2}[/tex])[tex]^\circ[/tex] =([tex]\frac{135}{2}[/tex])[tex]^\circ[/tex] =67[tex]^\circ[/tex] 30'

[tex]\angle[/tex]A+[tex]\angle[/tex]B+[tex]\angle[/tex]C =180[tex]^\circ[/tex]
[tex]\angle[/tex]A +67[tex]^\circ[/tex]30' +90[tex]^\circ[/tex]= 180[tex]^\circ[/tex]
[tex]\angle[/tex]A =22[tex]^\circ[/tex]30'

The decision is from Bulgaria .