In a triangle ABC, what is the angle $\theta$ between the median AM and the bisector AD? I want a way to know the measure of that angle, given the lenghts of the sides and angles of the triangle. I tried solving the triangle, with the sine and cosine laws, and I arrived at $$ \theta = \frac{A}{2} - \arcsin\left(\frac{2\sqrt{(a/2)^2+b^2 -ab\cos C}} {a\sin C}\right). $$ I'm not even totally sure if it's correct and it's a bit unwieldy... I am wondering if there is a smarter aproach that gives a nicer formula. Thank you!

  • $\begingroup$ What is $\Gamma?$ $\endgroup$ – saulspatz Apr 29 '18 at 17:12
  • $\begingroup$ sorry! I mean angle C, I wasn't consistent in the notation $\endgroup$ – George Ntoulios Apr 29 '18 at 17:14
  • 1
    $\begingroup$ See this. $\endgroup$ – MalayTheDynamo Apr 29 '18 at 17:24
  • $\begingroup$ nice! with this the sqare root in the formula becomes $\sqrt{\frac{b^2+c^2}{2}-a^2}$ it is much better for sure $\endgroup$ – George Ntoulios Apr 29 '18 at 17:41

Let $A = 2\alpha$, $B = 2\beta$, $C = 2\gamma$. We can coordinatize, with $$A = (0,0) \qquad B = c\left(\cos\alpha,-\sin\alpha\right) \qquad C = b\left(\cos\alpha,\sin\alpha\right) \qquad M = \frac12(B+C)$$

Since bisector $\overline{AD}$ coincides with the $x$-axis, $\angle DAM$ is simply the direction angle of $M$. Perhaps the simplest representation is

$$\tan DAM = \frac{M_y}{M_x} = \frac{B_y+C_y}{B_x+C_x} = \frac{(b-c)\sin\alpha}{(b+c)\cos\alpha} \tag{$\star$}$$

The Law of Sines tells us that $b = d \sin B$ and $c = d \sin C$, where $d$ is the circumdiameter of $\triangle ABC$. With trig's prosthaphaeresis formulas, and recalling that $\alpha + \beta + \gamma = 90^\circ$, we can rewrite $(\star)$ as

$$\tan DAM = \frac{\sin^2\alpha \sin(\beta-\gamma)}{\cos^2\alpha\cos(\beta-\gamma)} = \tan^2\alpha \tan(\beta-\gamma) \tag{$\star\star$}$$

Alternatively, using the half-angle formulas, we have $$\tan DAM = \frac{(1-\cos A)\sqrt{1-\cos(B-C)}}{(1+\cos A)\sqrt{1+\cos(B-C)}}$$ which, with the help of the Law of Cosines, is manipulatable into this lengths-only form

$$\tan DAM = \frac{b-c}{b+c}\;\sqrt{\frac{\phantom{-}(a - b + c)(a + b - c)}{(-a + b + c) (a + b + c)}} \tag{$\star\star\star$}$$


By law of sines: $$ \frac{\sin CAM}{\sin BAM}=\frac{\sin C}{\sin B}=:\lambda. $$

Setting $\sin BAM=x$, and solving the resulting equation for cosine of the sum: $$ \sqrt{1-x^2}\cdot\sqrt{1-\lambda^2x^2}-x\cdot\lambda x=\cos A\\ $$ one obtains: $$ \sin BAM=\frac{\sin A}{\sqrt{1+\lambda^2+2\lambda\cos A}} =\frac{\sin A\sin B}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}} $$ and similarly $$ \sin CAM=\frac{\sin A\sin C}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}}. $$ Correspondingly: $$ \cos BAM=\frac{\sin B\cos A+\sin C}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}},\\ \cos CAM=\frac{\sin B+\sin C\cos A}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}}. $$

Writing: $$\small{ \cos(BAM-CAM)=\frac{(\sin B\cos A+\sin C)(\sin B+\sin C\cos A)+\sin B\sin C\sin^2A}{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}\\ =\frac{(\sin^2 B+\sin^2 C)\cos A+2\sin B\sin C}{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A} } $$ and noticing that $|\widehat{BAM}-\widehat{CAM}|$ is the doubled value of the angle in question one finally obtains: $$ \theta=\frac{1}{2}\arccos\frac{(\sin^2 B+\sin^2 C)\cos A+2\sin B\sin C}{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.