Geometric solution

Redefine point $A$ to satisfy $MA=MC$ and $\angle MAC= \angle MCA=10^{\circ}.$ So, now it suffices to prove that $\angle ABM=30^{\circ}$ or $AB=AC.$ Let $D$ be a point (below $BC$) such that $MB=MD, \angle BMD= 40^{\circ},$ and $MD\cap BC\equiv E.$ Denote $I$ as the incenter of $\Delta MEC. F\in BM: ME=MF, G\in BC :FE=FG.$ Clearly, $\Delta MDA \cong \Delta MBC \Rightarrow \angle MDA=20^{\circ}, \angle MAD=40^{\circ}, \angle CAD=50^{\circ} --(1)$

We have; $\angle MFE=70^{\circ}$ and $\angle MIE= 90^{\circ}+\frac{\angle MCE}{2} = 110^{\circ} \Rightarrow$ quad. $MIEF$ is cyclic. Also, since $ME$ bisects $\angle IMF \Rightarrow IE=EF=GF$ Now, note that $\Delta BFG\sim \Delta CEI$ by $AAS-$ similarity $\Rightarrow BF=CE$ But due to symmetry in isosceles $\Delta BMD, BF=ED$ Combining the two we get, $CE=ED \Rightarrow \angle MDC= 30^{\circ} \therefore,$ from result $(1) \angle CDA=50^{\circ}=\angle CAD \Rightarrow CA=CD$ Also, $\Delta MCD\cong\Delta MAB \Longrightarrow AB=CD=AC$. And, so done. $\boxed{Q.E.D}$

Have you tried an algebraic solution to this, involving all the unknown angles?

P.S. All the methods used still fall under the banner of "trigonometry", as ultimately we are studying triangles.

@Kazem Sepehrinia Well, I don't have a solution without trigonometry, but we can definitely solve it quickly by applying the Sine Rule to Triangles AMC and AMB and take advantage of the fact that AB = AC along with AM = AM (the common side).....!!

i found a solution with basic geometry and got the answer 10 ... is it correct?

please draw step by step with me... extend AB to meet AC at D.Again extend MD to reach a point E which MD=DE.So as you can see AM=AE and triangle EBM is equilateral. Take a point P in triangle MEB such that angle PBA=10.Thus,two triangles ABP & AMC are congruent.So we got that AP=AM=AE and angle MAP=80 and PB=CM. Draw a circle with radius AM.As you can see angle PAM=2*PEM=80>>>>PEM=40>>>>>BEP=PBE=20>>>>>PB=PE. PB=BE PM=PM(common)>>>>>>>>>>>>>>>>>>>>triangles MPE & BPM are congruent>>>>>>>angles EMP=PMB=30>>>>>>>>angle PAE=60(in circle) BM=Em So triangle PEA is also equilateral>>>>>EP=AM=BP=CM. after all sorry for my English.... :)

try. this.. or **else . .. draw a perpendicular from a on bc.. and now. daw. the same figure symmetric abt.. AP then mark the point on AP where lines intersect as Q .. then.. produce. BM. till it meets AC .. and then also drop a perpendicular from M on AC name it N .repeat same... symmetrically. on othr side.. then.. use. basic geometry..(properties of isosceles triangle. and.. exterior angle theorem and. solve fr all the angles.. .. you will realise that. angle ANM is equal to angle AQM. equal to 70 degress. then use again properties of similar triangle.. n find. 2x=180 -2(70)=> x=20 this. is entirely. geometry

ABC is a triangle with angle bisector AD,BE,and CF. If angle FDE = 90 degree then find the angle BAC?

How it is came? Send me the COMPLETE solution .In this way i would able to findmy mistake. Of Q1, ABC is a triangle with angle bisector AD,BE,and CF. If angle FDE = 90 degree then find the angle BAC

I have solve it

It is 10

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take a protector.. scale compass.. draw thefig. and measure