Problem of the Week
Problem C and Solution
Angled

Problem

Line segments $AB$ and $CD$ are parallel, with $AB$ above $CD$. Point $J$ lies above $AB$, and points $E$ and $F$ lie on $CD$, with $E$ to the left of $F$, so that $JE$ intersects $AB$ at $G$ and $JF$ intersects $AB$ at $H$.

If $\mathrm{\angle }CEG=110\mathrm{°}$ and $\mathrm{\angle }GHF=122\mathrm{°}$, determine the measure of $\mathrm{\angle }GJH$.

Solution

Solution 1

Since $JHF$ is a straight line, then $\mathrm{\angle }JHG=180\mathrm{°}-\mathrm{\angle }GHF=180\mathrm{°}-122\mathrm{°}=58\mathrm{°}.$ Since $AB$ and $CD$ are parallel, $\mathrm{\angle }AGJ=\mathrm{\angle }CEG=110\mathrm{°}.$ Since $AGH$ is a straight line, $\mathrm{\angle }JGH=180\mathrm{°}-\mathrm{\angle }AGJ=180\mathrm{°}-110\mathrm{°}=70\mathrm{°}.$ Since the three angles in a triangle add to $180\mathrm{°}$, then $$\mathrm{\angle }GJH=180\mathrm{°}-\mathrm{\angle }JGH-\mathrm{\angle }JHG=180\mathrm{°}-70\mathrm{°}-58\mathrm{°}=52\mathrm{°}.$$

Solution 2

Since $JHF$ is a straight line, then $\mathrm{\angle }JHG=180\mathrm{°}-\mathrm{\angle }GHF=180\mathrm{°}-122\mathrm{°}=58\mathrm{°}.$ Since $AB$ and $CD$ are parallel, $\mathrm{\angle }JFE=\mathrm{\angle }JHG=58\mathrm{°}.$ Since $CEF$ is a straight line, $\mathrm{\angle }JEF=180\mathrm{°}-\mathrm{\angle }CEG=180\mathrm{°}-110\mathrm{°}=70\mathrm{°}.$ Since the three angles in a triangle add to $180\mathrm{°}$, then $$\mathrm{\angle }GJH=\mathrm{\angle }EJF=180\mathrm{°}-\mathrm{\angle }JEF-\mathrm{\angle }JFE=180\mathrm{°}-70\mathrm{°}-58\mathrm{°}=52\mathrm{°}.$$

Note: Since $\mathrm{△}GJH$ and $\mathrm{△}EJF$ have all three angles in common, we can say that they are similar triangles. Similar triangles have properties that make them very useful in geometry problems.