# Solution to the [[Three Congruent Triangles]] _Posted on Oct 29, 2019_ +-- {.image} [[ThreeCongruentTriangles.png:pic]] > Three congruent triangles. The black line bisects the top angle. What angle does this line make with the horizontal? =-- ## Solution by [[Angle at the Centre is Twice the Angle at the Circumference]], [[Equilateral Triangles]], [[Angles in a Triangle]] +-- {.image} [[ThreeCongruentTrianglesCircled.png:pic]] =-- 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]] +-- {.image} [[ThreeCongruentTrianglesSimilar.png:pic]] =-- 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 $E \hat{H} A = 105^\circ$.