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Problem 1573

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Geometry Difficulty 5.7 Prove it India — Team Selection Test · India

Let $ABCDE$ be a convex pentagon in which $BC \parallel AE$ , $AB = BC + AE$ and $\angle ABC = \angle CDE$ . Let $M$ be the mid-point of $CE$ and let $O$ be the circumcentre of triangle $BCD$ . Suppose $\angle DMO = 90^\circ$ . Prove that $2\angle BDA = \angle CDE$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let the circum-circle of $BCD$ be $\Gamma$ and let the circle with diameter $OD$ be $\Gamma'$ . Let the mid-point of $OD$ be $O'$ (which is also the centre of $\Gamma'$ ). Now $OO'$ passes through $D$ and hence $\Gamma'$ and $\Gamma$ are tangent to each other at $D$ . Hence there is a homothety with centre $D$ taking $\Gamma'$ to $\Gamma$ , with ratio $2$ . Since $\angle OMD = 90^\circ$ , the point $M$ is on $\Gamma'$ . Let the image of $M$ under the homothety be $D'$ ; it lies on $\Gamma$ and $M$ is the midpoint of $DD'$ .

Consider the half-turn centred at $M$ taking $D$ to $D'$ . This also takes $C$ to $E$ as $M$ is the mid-point of $CE$ . Let $B'$ be the image of $B$ under this transformation. Since the half turn takes a line to another line parallel to it, the image of $BC$ is $B'E$ ; and $BC \parallel B'E$ . Thus $AE \parallel BC \parallel B'E$ . It follows that $A$ , $B'$ , $E$ are collinear. Observe that $A$ , $B$ lie on the same side of $CE$ . Since we have performed a rigid transformation preserving $C$ , $E$ , the images of $A$ , $B$ must also lie on the same side of $CE$ . This implies that $A$ and $B'$ lie on different sides of $CE$ . Thus $E$ lies between $A$ and $B'$ . Because of this, we have
$\begin{aligned} AB' = AE + EB' &= AE + BC \quad (\text{half-turn takes } BC \text{ to } B'E) \\ &= AB. \end{aligned}$
We also observe that the half-turn takes triangle $CD'B$ to $EDB'$ . Therefore $\angle CD'B = \angle EDB'$ . Thus we get
$\begin{aligned} \angle BDB' = \angle BDE + \angle EDB' &= \angle BDE + \angle CD'B = \angle BDE + \angle CDB \\ &= \angle CDE = \angle CBA = 180^\circ - \angle BAB', \end{aligned}$
since $BC \parallel AE$ . This shows that $B$ , $D$ , $B'$ , $A$ are concyclic. Since $AB = AB'$ , they subtend equal angle at $D$ . Thus $\angle ADB' = \angle BDA'$ . Thus
$2\angle BDA = \angle BDA + \angle ADB' = \angle BDA + \angle ADE + \angle EDB'.$
But observe $\angle EDB' = \angle CD'B = \angle CDB$ . Using this we obtain
$2\angle BDA = \angle BDA + \angle ADE + \angle CDB = \angle CDE.$
This completes the proof.

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