# Proof With Squares, Triangles, and Midpoints

Let $ABC$ be a triangle. We construct squares $ABST$ and $ACUV$ with centers $O_1$ and $O_2$, respectively, as shown. Let $M$ be the midpoint of $\overline{BC}$. (a) Prove that $\overline{BV}$ and $\overline{CT}$ are equal in length and perpendicular.

(b) Prove that $\overline{O_1 M}$ and $\overline{O_2 M}$ are equal in length and perpendicular.

I tried drawing diagonals of the squares through their centers but that didn't help much. I'm out of ideas and stuck. All solutions are highly appreciated!

• use $BV=BA+AV$,$CT=CA+AT$ and angle $BAV=$ angle $TAC$. – hamam_Abdallah Oct 19 '16 at 23:36

Perform a $90^{\circ}$ degree counterclockwise rotation around point $A$. Then Point $T$ is mapped to point $B$ and point $C$ is mapped to point $V$. Therefore triangle $ABV$ is the image of triangle $ATC$ under the rotation. Thus the two triangles are congruent and segment $TC$ is mapped to $BV$ under the rotation, which means that $TC = BV$ and $TC$ is orthogonal to $BV$. Notice that segment $MO_1$ is a midsegment of triangle $BCT$ because $M$ is the midpoint of $BC$ and $O_1$ is the midpoint of $BT$. Hence, $MO_1 = \frac{1}{2} TC$ and $MO_1$ is parallel to $TC$. Analogously, segment $MO_2$ is a midsegment of triangle $CBV$ because $M$ is the midpoint of $BC$ and $O_2$ is the midpoint of $CV$. Hence, $MO_2 = \frac{1}{2} TC$ and $MO_2$ is parallel to $BV$. Since $TC$ and $BV$ are equal and perpendicular, $$MO_1 = \frac{1}{2} TC = \frac{1}{2} BV = MO_2 \,\,\, \text{ and } \,\,\, MO_1 \text{ is perpendicular to } MO_2$$
Draw the triangles $\;\Delta CTA\;,\;\;\Delta BVA\;$ and prove this two are congruent triangles (hint: (1) do separate drawings of both triangles and write on them, (2) show that $\;\angle TAC=\angle VAB)$