Geometric proof for the half angle tangent

Question

Using the fact that the angle bisector of the below triangle splits the opposite side in the same proportion as the adjacents sides, I need to give a geometric proof of the half-angle tangent (tan(beta/2) = sin(beta)/(1 + cos(beta))).

This is what I've done so far:
I tried to write the sides "a" and "c" in terms of b1 and b2 using the Pythagoras Theorem and the relations of sin and cos. After a lot of manipulation, I ended up with tan(Beta/2) = (sen(Beta)* b2ˆ2 * cos(Beta))/(b1 + b2). But this is certainly wrong.
Any hints on how to proceed?

Narasimham · Answer

Any fraction can as per a rule of fractions be algebraically also written to form an identity taking sum/difference of numerator and denominator separately with or without a common multiplier. Using this rule
$$ frac{p}{q}=frac{r}{s}=frac{ap+br}{aq+bs}$$
$$
tan beta/2=frac{b_1}{a}=frac{b_2}{c}=frac{b_1+b_2}{a+c}=frac{b}{a+c}=frac{b/c}{a/c+1}=frac{sinbeta}{cosbeta+1}.
$$

user10354138 · Answer

Since $dfrac{b_1}{a}=dfrac{b_2}{c}$ and $b_1+b_2=b$, we have
$$
tan(beta/2)=frac{b_1}{a}=frac{b}{a+c}=frac{(b/c)}{(a/c)+1}=frac{sinbeta}{cosbeta+1}.
$$

N. S. · Answer

Hint
If $frac{b_1}{a}=frac{b_2}{c}$ then
$$frac{b_1}{a}=frac{b_2}{c}=frac{b_1+b_2}{a+c}$$

Geometric proof for the half angle tangent

3 Answers

Add your own answers!

Ask a Question