Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it Bulgaria

Problem:
Let $ABCD$ be a circumscribed quadrilateral. Find $\text{BCD}$ if $AC = BC$ , $AD = 5$ , $E = AC \cap BD$ , $BE = 12$ and $DE = 3$ .

Solution

Solution:
If the perpendicular bisector of $CD$ meets $BD$ at point $O$ , then
$\text{COD = 180 - 2 ODC = 180 - 2 BAC = ACB = ADO}$
and therefore $AD \parallel CO$ . Hence $\frac{OE}{3} = \frac{CO}{AD} = \frac{OE + 3}{5}$ , which implies that $OE = \frac{9}{2}$ . This shows that $O$ is the midpoint of $BD$ and then

$\text{BCD = 90}$ .

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