Solving geometry question by vector and pure geometry.

Question

In $\Delta ABC$, $\angle A=\frac{\pi}{6}$, $H$ is the orthocentre and $D$ is the mid point of $BC$.

Segment $HD$ is produced to $T$ such that $HD=DT$. Length $AT=\lambda BC$. Find $\lambda$.

If possible, I am looking for method by pure geometry and vector both.

Let the circumcenter of $\Delta ABC$ be origin $(0)$

Position vector (P.V.) of $A,B,C$ be $a,b,c$ respectively.

$\implies$ P.V. of orthocentre $ H =\vec{a}+\vec{b}+\vec{c}$.

P.V. of $ D= \dfrac{\vec{b}+\vec{c}}{2}$.

Since $D$ is mid point of $H$ and $T$ $\implies \dfrac{\vec{b}+\vec{c}}{2}=\dfrac{(\vec{a}+\vec{b}+\vec{c})+\vec{t}}{2} \Longleftrightarrow \vec{t}=-\vec{a}$.

where $\vec{t}=$ P.V of $T$, $\vec{AT}=-2\vec{a}$

Edit: By the help from comment of user @Aretino, we see that $\vec{AO}$ and $\vec{OT}$ are parallel and same magnitude $\Longrightarrow$ length $AT$ is diameter of the circumcircle.

And by extended sine theorem, $BC =2R\sin(\pi/6)=R$.Hence $\lambda$ is 2.Now I am interested in pure geometrical proof?

Answer 1

Since $\angle BOC = 2\angle BAC = 60$ so $BCO$ is equilateral, so $BC = OB = AO =R$.

• Since $T$ is reflection over a midpoint of a segment $BC$ we see that $T\in (ABC)$
• Reflection $H'$ over a segment $BC$ is also on $(ABC)$ and since $ED||H'T$ we see that $\angle TH'A = 90^{\circ}$ so $AT = 2R$

Conclusion $\lambda = 2$.