# Whats is the relation between $AM$ and $BC$ in the question below?

For reference: (exact copy of the question) In a right triangle ABC, straight at $$"B"$$, the median $$AM$$ is drawn. If the measurement of angle $$AMB$$ is twice the measurement of angle $$A$$, calculate: $$\frac{AM}{BC}$$

My progress..By trigonometry I decided, I would like the resolution by geometry, if possible..

$$\triangle ABM: tan2\alpha=\frac{h}{x}(1)\\ \triangle ABC:tan\alpha=\frac{2x}{h}(2)\\ (1)x(2):tan2\alpha \cdot tan\alpha=2\\ \frac{2tan\alpha}{1−tan^2\alpha}\cdot tan⁡α=2\implies tg \alpha = \frac{\sqrt2}{2}(3)\\ (3)in(2):\\ \frac{\sqrt2}{2}=\frac{2x}{h}\implies h = 2\sqrt2x\\ (T.Pit)ABM:\\ x^2+(2\sqrt2x)^2=AM^2 \implies 3x=AM\\ \therefore \frac{AM}{BC} =\frac{3x}{2x}=\frac{3}{2}$$

I will solve it one of the ways and show another way to go about it but leave it for you to fill the details.

First way -

Say point $$D$$ is reflection of $$C$$ about $$AB$$ and $$N$$ is reflection of $$M$$ about $$AB$$.

We know, $$\angle BAM = 90^\circ - 2 \alpha$$

So, $$\angle DAM = \alpha + (90^\circ - 2 \alpha) = 90^\circ - \alpha$$

Also, $$\angle ADM = \angle ACB = 90^\circ - \alpha$$

And it follows that $$\triangle AMD$$ is isosceles with $$AM = DM = 3 x$$

So, $$\displaystyle \frac{AM}{BC} = \frac{3 x}{2 x} = \frac{3}{2}$$

Second way -

Midpoint of $$AC$$ (point $$N$$) is circumcenter of the right triangle $$\triangle ABC$$. Intersection $$G$$ of $$AM$$ and $$BN$$ is the centroid, so $$AG:GM = ?$$. Now can you show in $$\triangle BGM$$, $$BM = GM$$ and can you finish the proof?

• $\triangle BNC (isosceles): \angle NCB =\angle CBN = 90-\alpha\implies \angle BGM = 90-\alpha \therefore \triangle BGM (isosceles)\\ \therefore BM = GM=x\\ but ~AG = 2GM = 2x \implies AM = 3x \\ \therefore \frac{AM}{BC}=\frac{3}{2}$ Oct 5, 2021 at 19:47

Construct bisector of $$\angle AMB$$. Then draw a perpendicular from $$N$$ to $$AM$$. It follows that $$\triangle MON \cong \triangle MBN \implies MO=x$$.
Also, $$\triangle MBN \sim \triangle ABC \implies \frac{BN}{x}=\frac{2x}{h} \implies BN=\frac{2x^2}{h}$$
Finally, $$\triangle AON \sim \triangle ABM \implies \frac{AO}{h}=\frac{ON}{x}=\frac{2x^2}{xh} \implies AO=2x \implies AM=3x$$

• great view ..thankful for the help Oct 4, 2021 at 21:02