ABC is a triangle with a right angle at C, if the median of the side $c$ is the geometric mean of sides $a$ and $b$

Question

$ABC$ is a triangle with a right angle at $C$, if the median of the side $c$ is the geometric mean of sides $a$ and $b$, prove that one of the acute angles of $ABC$ is $15^o$

I've Pythagoras theorem but got nothing.

Answer 1

Let $CA=b$, $CB=a$, $AB=c$, $CD$ be a median of $\Delta ANC$ and $E$ placed on line $CD$ such that $C$ is a midpoint of $ED$.

Also, let $\Phi$ be a circumcircle of $\Delta AED$ and $AC\cap \Phi=\{A,F\}$.

Since $CD^2=CD\cdot EC=AC\cdot CF$ and from the given $CD^2=AC\cdot BC$, we get $CF=CB$.

In another hand, $\measuredangle A=\measuredangle E$.

Thus by law of sines for $\Delta ECF$ we obtain: $$\frac{\sin\measuredangle F}{EC}=\frac{\sin\measuredangle E}{CF}$$ or $$\frac{\sin\measuredangle F}{\frac{c}{2}}=\frac{\sin\measuredangle A}{a}$$ or $$\frac{\sin\measuredangle F}{\frac{c}{2}}=\frac{\frac{a}{c}}{a}$$ or $$\sin\measuredangle F=\frac{1}{2}.$$

But $$\measuredangle B=\frac{1}{2}\measuredangle EDA=\frac{1}{2}\measuredangle F$$ and since $\measuredangle F=30^{\circ}$ or $\measuredangle F=150^{\circ}$, we get the answer: $\measuredangle B=15^{\circ}$ or $\measuredangle B=75^{\circ}$.

Answer 2

from the Problem is given we have $$R=\sqrt{ab}$$ and $$\sin(\alpha)=\frac{a}{2\sqrt{ab}}=\frac{1}{2}\sqrt{\frac{a}{b}}$$ from $$\frac{c}{2}=\sqrt{ab}$$ we get by squaring $$a^2+b^2=4ab$$ dividing by $$ab\ne 0$$ we get $$\frac{a}{b}+\frac{b}{a}=4$$ Setting $$t=\frac{a}{b}$$ we get the quadratic equation $$t^2-4t+1=0$$ and solving this we have $$t_{1,2}=2\pm\sqrt{3}$$ can you finish from here? Setting $2-\sqrt{3}$ in the formula you will get $$\sin(\alpha)=\frac{1}{2}\sqrt{2-\sqrt{3}}$$ this is the formula for $$\sin(15^{\circ})$$

Answer 3

Property: In a right triangle the median is equal to a half of the hypotenuse.

The median is geometric mean of the legs: $$m_c=\sqrt{ab} \Rightarrow \frac{c}{2}=\sqrt{ab} \Rightarrow c=2\sqrt{ab}.$$

Find the sine of the acute angle $A$: $$\sin{A}=\frac{a}{c}=\frac{a}{2\sqrt{ab}}=\frac{1}{2}\sqrt{\frac{a}{b}} \Rightarrow $$ $$\sin^2{A}=\frac{1}{4}\cdot \frac{a}{b}=\frac{1}{4}\tan{A}=\frac{\sin{A}}{4\cos{A}} \Rightarrow$$ $$2\sin{A}\cos{A}=\frac{1}{2} \Rightarrow \sin{2A}=\frac{1}{2} \Rightarrow 2A=30^o or 150^o \Rightarrow A=15^o or 75^o \Rightarrow B=75^o or 15^o.$$