Let $ A'$ the point such that $ BACA'$ is a paralellogram. Let $ M' \in A'C$ such that $ D$ is the midpoint of $ MM'$, let $ N' \in A'B$ such that $ D$ is the midpoint of $ NN'$. Define $ b = BM = M'C$, $ c = CN = BN'$, $ m = AM = A'M'$, $ n = AN = A'N'$, $ p=MD=DM'$, $ q=ND=DN'$, $ \angle \alpha = \angle MAN = \angle MDN$. By the Law of Cosines: in $ MAND$ : $ m^2 + n^2 - 2mn\cdot cos\alpha = p^2 + q^2 - 2pq\cdot cos\alpha$ in $ BMDN'$: $ b^2 + c^2 + 2bc\cdot cos\alpha = p^2 + q^2 + 2pq\cdot cos\alpha$ Suposse that $ \alpha \neq 90^\circ$, then $ cos\alpha\neq 0$ since $ m^2 + n^2 = b^2 + c^2$ we get $ bc + mn = 2pq$... (1) Since $ [BACA'] = 2[BMN'] + 2[MAN] + [MNM'N']$ we get $ (b + m)(c + n)\cdot sin\alpha = (bc + mn + 2pq)\cdot sin\alpha$ we get $ bn + cm = 2pq$... (2). By (1) and (2) we have $ (b - m)(c - n) = 0$, contradiction!

Nice solution Tipe.

Valentin Virgil Nicula wrote: Let $ ABC$ be a triangle and $ D$ be the midpoint of the side $ [BC]$ . Exist $ M\in (AB)$ , $ N\in (AC)$ so that $ AM^2 + AN^2 = BM^2 + CN^2\ \ \wedge\ \ \widehat {MDN}\equiv\widehat {BAC}$ . Prove that $ AB\perp AC\ \ \vee\ \ AB=AC$ . Virgil Nicula wrote: Lemma (trigonometric form of the Ptolemeus' relation in a cyclic quadrilateral). Let $ ABC$ be a triangle. Then for any $ \phi\in R$ exists the identity $ a\cdot\sin\phi + b\cdot\sin (C - \phi ) = c\sin (B + \phi )$ . Particular cases : $ \{\begin{array}{ccccc} \blacktriangleright & \phi : = 0 & \implies & b\cdot\sin C = c\cdot\sin B & \mathrm {(Sinus'\ theorem)} \\ \\ \blacktriangleright & \phi : = \frac {\pi}{2} - B & \implies & a\cdot\cos B + b\cdot\cos A = c & \mathrm {(Cosinus'\ theorem)}\end{array}$ Proof. Denote $ \{\begin{array}{c} BM = m\ ,\ CN = n \\ \\ m(\widehat {BDM}) = x\ ,\ m(\widehat {CDN}) = y\end{array}$ . Thus, $ \{\begin{array}{c} AM = c - m\ ,\ AN = b - y \\ \\ x + y + A = 180^{\circ}\end{array}$ and $ AM^2 + AN^2 = BM^2 + CN^2$ $ \Longleftrightarrow$ $ 2(mc + nb) = b^2 + c^2$ $ \Longleftrightarrow$ $ \boxed {\ c(2m - c) + b(2n - b) = 0\ }\ \ (1)$ . Apply the Sinus' theorem in the triangles $ BDM$ and $ CDN$ : $ \{\begin{array}{ccccc} \frac {BM}{\sin\widehat {BDM}} = \frac {DB}{\sin\widehat {DMB}} & \implies & \frac {m}{\sin x} = \frac {a}{2\sin (B + x)} & \implies & 2m = \frac {a\sin x}{\sin (B + x)} \\ \\ \frac {CN}{\sin\widehat {CDN}} = \frac {DC}{\sin\widehat {DNC}} & \implies & \frac {n}{\sin y} = \frac {a}{2\sin (C + y)} & \implies & 2n = \frac {a\sin y}{\sin (C + y)}\end{array}$ . From the relation $ (1)$ obtain : $ \frac {c}{\sin (B + x)}\cdot[a\sin x - c\sin (B + x)] + \frac {b}{\sin (C + y)}\cdot[a\sin y - b\sin (C + y)] = 0$ . Using the upper lemma obtain $ \frac {c}{\sin (B + x)}\cdot b\sin (x - C) + \frac {b}{\sin (C + y)}\cdot c\sin (y - B) = 0$ $ \Longleftrightarrow$ $ \sin (x - C)\sin (C + y) + \sin (B + x)\sin (y - B) = 0$ $ \Longleftrightarrow$ $ \cos(2C + y - x) - \cos(x + y) + \cos (2B + x - y) - \cos (x + y) = 0$ $ \Longleftrightarrow$ $ 2\cos (x + y) = \cos (2C + y - x) + \cos (2B + x - y)$ $ \Longleftrightarrow$ $ \cos (x + y) = \cos (B + C)\cos (x - y)$ $ \Longleftrightarrow$ $ \cos A = \cos A\cos (x - y)$ $ \Longleftrightarrow$ $ A = 90^{\circ}\ \ \vee\ \ x = y = 90^{\circ} - \frac A2$ . Prove easily that $ x=y\ \Longleftrightarrow\ AB=AC$ and in this case $ [MN]$ is the $ A$ - middleline of $ \triangle ABC$ , i.e. $ MA=MB$ , $ NA=NC$ .

My Solution: Let $ E, F$ be midpoints of $ AC, AB$ respectively. From the condition of the problem we have $ AM^2 - BM^2 = CN^2 - AN^2$ which is equivalent to $ AB(AM-BM) = AC(CN-AN)$. Since $ M, N$ are different from $ E, F$ we have $ \dfrac{AB}{AC} = \dfrac{DE}{DF} = \dfrac{CN-AN}{AM-BM} = \dfrac{2NE}{2MF} = \dfrac{NE}{MF}$ From this we derive that $ \triangle NED \sim \triangle MFD$ and this leads to $ \angle{ABC}=\angle{DEC}=\angle{MFD}=\angle{NED}$ But $ \angle{DEC}+\angle{NED} = 180^{\circ}$ We therefore have $ \angle{BAC}=90^{\circ}$.

mathangel wrote: My Solution: Let $ E, F$ be midpoints of $ AC, AB$ respectively. From the condition of the problem we have $ AM^2 - BM^2 = CN^2 - AN^2$ which is equivalent to $ AB(AM - BM) = AC(CN - AN)$. Since $ M, N$ are different from $ E, F$ we have $ \dfrac{AB}{AC} = \dfrac{DE}{DF} = \dfrac{CN - AN}{AM - BM} = \dfrac{2NE}{2MF} = \dfrac{NE}{MF}$ From this we derive that $ \triangle NED \sim \triangle MFD$ and this leads to $ \angle{ABC} = \angle{DEC} = \angle{MFD} = \angle{NED}$ But $ \angle{DEC} + \angle{NED} = 180^{\circ}$ We therefore have $ \angle{BAC} = 90^{\circ}$. Why $ \triangle NED \sim \triangle MFD$ ? Thanks

Because <MDF=<NDE

I really liked the problem, despite the problem somewhat hinting towards the use of midpoints. Let $L, K$ be the midpoints of the sides $AB, AC$, respectively. WLOG, let $AM > AL$, $AN < AK$. We'll prove that, $$ AB/AC = NK/LM $$which will give that $NK/LM = KD/ LD$ or $NK/KD = LM/ LD$, which we can then use law of sines on $\triangle NKD, \triangle MLD$. By our condition, $AM^2 - BM^2 = AB(AM - BM) = AC(CN - AN) = CN^2 - AN^2$, or since $AM - BM = 2LM$ and $CN - AN = 2NK$, we get our claim. Let $\angle NDK = \angle MDL = a$, $\angle BLD = \angle DKC = b$. By Law of Sines on $\triangle NKD$ and $\triangle MLD$, we get $$NK/KD = \sin a/ \sin {(b-a)} = \sin a/ \sin (a + b) = LM/LD$$ or $\sin (b-a) = \sin (a+b)$. Two cases: $1)$ $b - a = a + b$ Then $a = 0$, contradiction. $2)$ $b - a = 180^{\circ} - a - b$ Then $b = 90^{\circ}$, which is what we needed to prove.