For $\Delta ABC$ right-angled triangle at $A$ have $AB=AC$. Calculate the ratio of $MA: MB: MC$

Question

For $\Delta ABC$ right-angled triangle at $A$ have $AB=AC$, uppose in the triangle that the point M satisfies $\widehat{MBA}=\widehat{MAC}=\widehat{MCB}$. Calculate the ratio of $MA: MB: MC$

Let D,E sequence are center circumsribed circle ABM, CAM

$\widehat{MAC} =\widehat{MBA}$

=>AC is tangent of the circle (D)

=>CA_|_ DA=> $D\in AB$

So $DA =DB$ =>D midpoint of AB

Have $\widehat{MCB} =\widehat{MAC}$

=>BC is tangent of d and midperpendicular of AC

(D) cut BC at F =>AF_|_BC and F midpoint BC

Have $\widehat{MBA} =\widehat{MCB}$ and $\widehat{MAB} =\widehat{MFC}$ (AMFB cyclic quadrilateral)

=>$\triangle MAB \sim\triangle MFC$=>$\frac{MA}{MB} =\frac{MF}{MC}(1)$

Have $FA=\frac{BC}{2}=FC$

=>FE is midperpendicular AC

FE cut AC at G. Have $\widehat{AGF} =\widehat{CGE}, \widehat{AFG} =\widehat{CEG}$; $GA;GC$

=>$\triangle AGF =\triangle CGE$

............. blah blah

$$\Rightarrow MA:MB:MC=1:2:\sqrt{2}$$

This is my try, it's very long and difficult, because i need new method

Answer 1

Forget the circles.

Scale triangle $ABC$ so that $AB = 1,\;AC = 1,\;BC=\sqrt{2}$.

Let $\;u = MA,\;v = MB,\;w = MC$.

Let $\theta$ be the common degree measure of angles $MBA,\;MAC,\;MCB$.

Chasing some angles, we get . . . $$\angle MAC = \theta,\;\angle ACM = 45-\theta,\;\angle CMA=135$$ $$\angle MCB = \theta,\;\angle CBM = 45-\theta,\;\angle BMC=135$$ Thus, triangles $MCB$ and $MAC$ are similar, hence, equating ratios of corresponding sides, we get $$\frac{\sqrt{2}}{1}=\frac{v}{w}=\frac{w}{u}$$ It follows that $$u\;\colon v\;\colon w = 1\;\colon 2\;\colon \sqrt{2}$$

Answer 2

Let $\,\widehat{MBA}=\alpha=\widehat{MCB}=\widehat{MAC}\,$, then:

$$\require{cancel} \widehat{AMB}= \pi - \widehat{MBA}-\widehat{MAB}= \pi - \widehat{MBA}-\left(\frac{\pi}{2}-\widehat{MAC}\right)=\pi-\bcancel{\alpha}-\left(\frac{\pi}{2}-\bcancel{\alpha}\right)=\frac{\pi}{2} $$

Similarly, $\,\widehat{BMC}=\pi-\left(\cfrac{\pi}{4}-\bcancel{\alpha}\right)-\bcancel{\alpha}=\cfrac{3\pi}{4}\,$, so $\,\widehat{CMA}=2 \pi - \widehat{AMB} - \widehat{BMC}=\cfrac{3\pi}{4}\,$.

Then, by the law of sines:

• $\;\triangle MAB\,$: $\;\;MA / \sin(\alpha) = AB / \sin(\pi/2) = AB$

• $\;\triangle MBC\,$: $\;\;MB / \sin(\alpha) = BC / \sin(3\pi/4) = \sqrt{2} \,BC = 2 \,AB$

• $\;\triangle MCA\,$: $\;\;MC / \sin(\alpha) = AC / \sin(3\pi/4) = \sqrt{2} \,AC = \sqrt{2} \,AB$

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[ EDIT  (+fix) ]  May also be worth noting that the above gives $\alpha=\arctan(1/2)$, so the point $\,M\,$ is uniquely determined by the conditions, and the construction can be fully solved. Asking for just the ratios of the distances to the vertices looks more like a diversion.