sqing wrote: Let $ABC$ be a triangle , $AD$ an altitude and $AE$ a median . Assume $B,D,E,C$ lie in that order on the line $BC$ . Suppose the incentre of triangle $ABE$ lies on $AD$ and he incentre of triangle $ADC$ lies on $AE$ . Find ,with proof ,the angles of triangle $ABC$ . Let $I_1, I_2$ be the incircles of $ABE, ADC$ respectively. $ABE$ is isosceles and $B\widehat AD = D\widehat AE = E\widehat AC = \theta $. Let $\widehat B = 2\omega ,\widehat C = 2\varphi $ From triangles $ABD, ADC$ we have $\boxed{2\omega+\theta=90^0}$ $(1)$ and $\boxed{\varphi+\theta=45^0}$ $(2)$ $B\widehat {{I_1}}E = D\widehat {{I_2}}C = {90^0} + \theta \Leftrightarrow {180^0} - 2\omega = {180^0} - {45^0} - \varphi \Leftrightarrow $ $\boxed{2\omega = {45^0} + \varphi }$ $(3)$ From $(1), (2), (3)$ we get $\theta=\omega=30^0$ and $\varphi=15^0$. Hence: $\boxed{\widehat{A}=90^0, \widehat{B}=60^0, \widehat{C}=30^0}$

george_54 wrote: sqing wrote: Let $ABC$ be a triangle , $AD$ an altitude and $AE$ a median . Assume $B,D,E,C$ lie in that order on the line $BC$ . Suppose the incentre of triangle $ABE$ lies on $AD$ and he incentre of triangle $ADC$ lies on $AE$ . Find ,with proof ,the angles of triangle $ABC$ . Let $I_1, I_2$ be the incircles of $ABE, ADC$ respectively. $ABE$ is isosceles and $B\widehat AD = D\widehat AE = E\widehat AC = \theta $. Let $\widehat B = 2\omega ,\widehat C = 2\varphi $ From triangles $ABD, ADC$ we have $\boxed{2\omega+\theta=90^0}$ $(1)$ and $\boxed{\varphi+\theta=45^0}$ $(2)$ $B\widehat {{I_1}}E = D\widehat {{I_2}}C = {90^0} + \theta \Leftrightarrow {180^0} - 2\omega = {180^0} - {45^0} - \varphi \Leftrightarrow $ $\boxed{2\omega = {45^0} + \varphi }$ $(3)$ From $(1), (2), (3)$ we get $\theta=\omega=30^0$ and $\varphi=15^0$. Hence: $\boxed{\widehat{A}=90^0, \widehat{B}=60^0, \widehat{C}=30^0}$ Wonderful. Thanks.

george_54 wrote: sqing wrote: Let $ABC$ be a triangle , $AD$ an altitude and $AE$ a median . Assume $B,D,E,C$ lie in that order on the line $BC$ . Suppose the incentre of triangle $ABE$ lies on $AD$ and he incentre of triangle $ADC$ lies on $AE$ . Find ,with proof ,the angles of triangle $ABC$ . Let $I_1, I_2$ be the incircles of $ABE, ADC$ respectively. $ABE$ is isosceles and $B\widehat AD = D\widehat AE = E\widehat AC = \theta $. Let $\widehat B = 2\omega ,\widehat C = 2\varphi $ From triangles $ABD, ADC$ we have $\boxed{2\omega+\theta=90^0}$ $(1)$ and $\boxed{\varphi+\theta=45^0}$ $(2)$ $B\widehat {{I_1}}E = D\widehat {{I_2}}C = {90^0} + \theta \Leftrightarrow {180^0} - 2\omega = {180^0} - {45^0} - \varphi \Leftrightarrow $ $\boxed{2\omega = {45^0} + \varphi }$ $(3)$ From $(1), (2), (3)$ we get $\theta=\omega=30^0$ and $\varphi=15^0$. Hence: $\boxed{\widehat{A}=90^0, \widehat{B}=60^0, \widehat{C}=30^0}$ This solution is wrong as the system of equations (1), (2), (3) do not have an unique solution(i.e. we cannot obtain the values of those angles, so the claim in this solution is wrong). You might think that they must have a unique solution since there are 3 variables and we have 3 equation, but in reality, equations (1), (2) imply equation (3). So if we see logically, we have only two equations (and there are three variables). So the thing claimed by that person in the end (who was the author of this solution) is wrong (according to me). Check the official solution in order to understand the correct solution to this problem

uhhhh we can get the same claims by noting that since $AH$ and $AO$ are isogonal wrt $\angle{A}$, we have that $AO$ bisects $BC$. But the projection from $O$ to $BC$ is the midpoint of $BC$, so $O$ has to lie on $BC$ => $\angle{A} = 90^\circ$. Similarly we get that since $\angle{BAE} = 2\angle{ACB} = \angle{AEB}$ and $AE = BE$, we have that $ABE$ is equilateral so the triangle is 30-60-90.