Question

22 given: ∠e = 70°; \\(\overrightarrow{jh}\\) and \\(\overrightarrow{fh}\\) bisect the exterior angles of △jef at j and f.
a find m∠h.
b can you find a formula that expresses m∠h in terms of m∠e?

Explanation:

Step1: Define exterior angles

Let exterior angles at J and F be $\angle EJK$ and $\angle EFG$.
$\angle EJK = 180^\circ - \angle EJF$, $\angle EFG = 180^\circ - \angle EFJ$.

Step2: Use angle bisector property

$\angle HJF = \frac{1}{2}\angle EJK = \frac{1}{2}(180^\circ - \angle EJF) = 90^\circ - \frac{1}{2}\angle EJF$
$\angle HFJ = \frac{1}{2}\angle EFG = \frac{1}{2}(180^\circ - \angle EFJ) = 90^\circ - \frac{1}{2}\angle EFJ$

Step3: Sum angles in $\triangle JEF$

$\angle E + \angle EJF + \angle EFJ = 180^\circ \Rightarrow \angle EJF + \angle EFJ = 180^\circ - 70^\circ = 110^\circ$

Step4: Calculate $\angle HJF + \angle HFJ$

$\angle HJF + \angle HFJ = (90^\circ + 90^\circ) - \frac{1}{2}(\angle EJF + \angle EFJ) = 180^\circ - \frac{1}{2}(110^\circ) = 125^\circ$

Step5: Find $\angle H$

$\angle H = 180^\circ - (\angle HJF + \angle HFJ) = 180^\circ - 125^\circ = 55^\circ$

Step6: Derive general formula

Let $\angle E = x$. Then $\angle EJF + \angle EFJ = 180^\circ - x$.
$\angle HJF + \angle HFJ = 180^\circ - \frac{1}{2}(180^\circ - x) = 90^\circ + \frac{x}{2}$.
$\angle H = 180^\circ - (90^\circ + \frac{x}{2}) = 90^\circ - \frac{x}{2}$.

Answer:

a. $55^\circ$
b. $m\angle H = 90^\circ - \frac{1}{2}m\angle E$