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three congruent triangles solution

Solution to the Three Congruent Triangles

Posted on Oct 29, 2019

Solution by Angle at the Centre is Twice the Angle at the Circumference, Equilateral Triangles, Angles in a Triangle

Let the points be labelled as above. Since the triangles are congruent, and $A G C$ is a straight line (as an edge of the upper triangle), angles $C \hat{G} E$ and $E \hat{G} A$ are both right-angles. Then angle $C \hat{B} A$ is also a right-angle, and so since $E B$ bisects angle $C \hat{B} A$, angle $E \hat{B} A = 45^\circ$. Also, as the triangles are congruent, $E G$, $G A$, and $A B$ all have the same length.

Draw a circle with centre $G$ that passes through $E$. Then it must also pass through $A$. Since angle $E \hat{B} A$ is half of angle $E \hat{G} A$, by the converse to the result that the angle at the centre is twice the angle at the circumference, $B$ must lie on the circle. Therefore, $B G$ has the same length as $A G$.

Since also $A B$ has that same length, triangle $A B G$ is equilateral and so angle $B \hat{A} H = 60^\circ$. Then since angle $H \hat{B} A = 45^\circ$, angle $B \hat{H} A = 180^\circ - 60^\circ - 45^\circ = 75^\circ$ and so angle $E \hat{H} A = 105^\circ$.

Solution by Similar Triangles, Isosceles Triangles, and Vertically Opposite Angles

In the picture above, $J$ is the midpoint of $A E$, and $I$ is where $G J$ meets $E B$. As in the previous argument, $G E$, $G A$, and $A B$ all have the same length.

Since $G J$ bisects angle $E \hat{G} A$, angle $I \hat{G} H$ is equal to angle $H \hat{B} A$. Angles $G \hat{H} I$ and $B \hat{H} A$ are vertically opposite. Therefore triangles $G H I$ and $B H A$ are similar, with vertices corresponding as listed.

This means that the ratios of the lengths of $H I : H A$ and of $G H : H B$ are the same. Since also angles $I \hat{H} A$ and $G \hat{H} B$ are equal, triangles $G H B$ and $I H A$ are also similar. Therefore, angle $G \hat{B} H$ is equal to angle $H \hat{A} I$. Then since both triangles $E G A$ and $E I A$ are isosceles, angles $G \hat{A} I$ and $G \hat{E} I$ are equal. Therefore, angles $G \hat{E} H$ and $G \hat{B} H$ are also equal. This means that triangle $E G B$ is also isosceles, and so $G B$ has the same length as $G E$.

Therefore, triangle $G B A$ is equilateral. Then, as above, angle $B \hat{A} H = 60^\circ$. Since angle $H \hat{B} A = 45^\circ$, angle $B \hat{H} A = 180^\circ - 60^\circ - 45^\circ = 75^\circ$ and so angle $E \hat{H} A = 105^\circ$.