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My problem is the following: given that $ABCDE$ is a convex pentagon such that $AB=BC$, $CD=DE$, $M$ is middle point of side $EA$ and the angles $\widehat{ABC}=\widehat{CDE}=90°$, find the measure of the angle $\widehat{BMD}$.

I think that $\widehat{BMD}=90°$, too, and it is possible to prove it through the cosine theorem applied to $ACE$ and the Pythagorean theorem applied to $BMD$. However, this approach is hard and tedious, I wonder if an easier solution exists.

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2 Answers

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It is a right angle.

This can be stated in the following way: if $ABC$ is a triangle, $O_B,O_C$ are the center of the squares (externally) built on $AC$ and $AB$ and $M$ is the midpoint of $BC$, then $MO_B \perp MO_C $.

I believe this is also known as Van Aubel's theorem.

Proof: embed the construction in $\mathbb{C}$ and assume without loss of generality $B=-1,C=1$.

We have $M=0$ and $O_C = (A-B)\cdot\frac{1+i}{2}+B$ and $O_B=(A-C)\cdot\frac{1-i}{2}+C$, so:

$$O_C = \frac{1+i}{2} A +\frac{i-1}{2},\qquad O_B = \frac{1-i}{2}A+\frac{1+i}{2}$$ and: $$ O_C = i\cdot O_B. $$ This proves $MO_C\perp MO_B$ and $MO_B=MO_C,$ too.

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Let $C={\bf 0}$, put $A=2{\bf a}$, $E=2{\bf e}$, and denote by $*{\bf x}$ the vector ${\bf x}$ turned $90^\circ$ counterclockwise. Then $$M={\bf a}+{\bf e}, \quad D={\bf e}-*{\bf e}, \quad B={\bf a}+*{\bf a}\ .$$

It follows that $$\overrightarrow{BM}=({\bf a}+{\bf e})-({\bf a}+*{\bf a})={\bf e}-*{\bf a}$$ and then $$\overrightarrow{DM}=({\bf a}+{\bf e})-({\bf e}-*{\bf e})={\bf a}+*{\bf e}=*({\bf e}-*{\bf a})=*\overrightarrow{BM}\ ,$$ which shows that $|BM|=|DM|$ as well.

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