Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Prove it Estonia

The diagonals of the quadrilateral ABDE meet at C. The segments AB and CE are of equal length 8 cm, and the segments AE and CD are also of equal length. The perimeter of the triangle CDE is 35 cm. Given that $\angle BAC = \angle AEC$ , find the perimeter of the pentagon ABCDE.

Solution

Using the condition of the problem, we get
$\angle BAE = \angle BAC + \angle CAE = \angle AEC + \angle CAE$
(Fig. 22). From the triangle $ACE$ we get $\angle AEC + \angle CAE = \angle DCE$ . Thus $\angle BAE = \angle DCE$ . At the same time, $AE = CD$ and $AB = CE$ . Consequently, the triangles $AEB$ and $CDE$ are equal, hence the perimeter of the triangle $AEB$ is $35$ cm.

Fig. 22

Now we get
$\begin{aligned} EA+AB+BC+CD+DE &= (EA+AB+BE-CE) + (CD+DE+EC-CE) \\ &= (EA+AB+BE) + (CD+DE+EC) - 2CE. \end{aligned}$
Since $EA + AB + BE = CD + DE + EC = 35 \text{ cm}$ and $CE = 8 \text{ cm}$ , the perimeter of the pentagon $ABCDE$ is $2 \cdot 35 \text{ cm} - 2 \cdot 8 \text{ cm} = 54 \text{ cm}$ .

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