Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 5.1 AIME, harder Prove it Ukraine

There is an equilateral trapezoid with bases $BC$ and $AD$ and known angles: $\angle BDC = 10^\circ$ and $\angle BDA = 70^\circ$ . Prove that the following equality holds true: $AD^2 = BC(AD + AB)$ .

Solution

It is clear that $\angle AMD = 20^\circ$ , $\angle DBM = 150^\circ$ . Let's construct equilateral triangle $\triangle KMD$ , then points $M$ , $B$ , $D$ are on the circle centered at $K$ (fig. 8.76). After that $KM = KB$ , $\angle KMB = 80^\circ$ , from where $\angle MBK = 80^\circ$ , but where $\angle MBC = 80^\circ$ as well, which implies that points $B$ , $C$ , $K$ are on the same line.

Then $\triangle AMD = \triangle BKM$ , as isosceles with equal sides and angles at the base. Therefore $MB = AD$ . From similarity of $\triangle MBC \sim \triangle MAD$ we have that $\frac{BC}{AD} = \frac{MB}{MA}$ $\Rightarrow$ $AD \cdot BM = BC \cdot MA$ , hence we obtain that

Fig. 45

$AD^2 = BC(MB + AB) = BC(AD + AB).$

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