Geometric proof for the half angle tangent

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))).

enter image description here

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?

Mathematics Asked by brazilian_student on November 12, 2021

3 Answers

3 Answers

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

$$ frac{p}{q}=frac{r}{s}=frac{ap+br}{aq+bs}$$ $$ tan beta/2=frac{b_1}{a}=frac{b_2}{c}=frac{b_1+b_2}{a+c}=frac{b}{a+c}=frac{b/c}{a/c+1}=frac{sinbeta}{cosbeta+1}. $$

Answered by Narasimham on November 12, 2021

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

Answered by user10354138 on November 12, 2021


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

Answered by N. S. on November 12, 2021

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