Geometric solution

I found a trigonometric solution to the following problem: Can you find a solution without using trigonometry? Note by Kazem Sepehrinia
10 months, 2 weeks ago

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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. $\boxed{Q.E.D}$

- 9 months, 3 weeks ago

Beautiful solution, thanks :)

- 5 months, 3 weeks ago

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.

- 10 months, 2 weeks ago

I used sin law several times and then a trigonometric equation comes out for the unknown MAC angle. Solving this equation gives MAC=10 degrees. But I wanted to use only basic geometric rules like parallel lines, isosceles triangles, sum of angles of a triangle. For example if we could prove AM=MC then problem is solved.

- 10 months, 2 weeks ago

You can solve it algebraically, as i said, in terms of the unknown angles, for which there are four. Fortunately, all the equations will be linear, so you will need to (hopefully) find four relations from the triangle that allow you to solve using linear algebraic methods.

To start you off, the angles around M add up to a full circle, so you can find a relation between two of the four unknown angles (as one of the angles is easily obtainable); this relation would namely be the sum.

- 10 months, 2 weeks ago

@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).....!!

- 10 months, 2 weeks ago

There are four unknown angles (as the fifth can be easily obtained), and four possible relationships involving the unknown which turn out to be all linear in the unknowns. This must mean that a linear algebraic solution can easily be obtained.

- 10 months, 2 weeks ago

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

- 10 months, 2 weeks ago

- 10 months, 2 weeks ago

You've got it right.. Please post your solution :)

- 10 months, 1 week ago

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.... :)

- 10 months, 1 week ago

Nice natural approach:)) There is a bit typo in the first line, it should be "CM" rather than "AB".

- 9 months, 3 weeks ago

I have also a new method

- 4 months, 1 week ago

I have proved it without congruence

- 4 months, 1 week ago

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

- 10 months, 1 week ago

Sorry, but the correct answer is 10 degrees.............

- 10 months, 1 week ago

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

- 8 months, 1 week ago

$120^{\circ}$ ---

- 8 months, 1 week ago

DUDE...........you beat me to it.......I was about to answer!!! XD

- 8 months, 1 week ago

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

- 8 months, 1 week ago

I have solve it

- 4 months, 1 week ago

It is 10

- 4 months, 1 week ago

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- 1 month ago

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- 1 month ago

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- 4 weeks, 1 day ago

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

- 10 months, 1 week ago