Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it Argentina

Let ABCDABCD be a convex quadrilateral that satisfies the following conditions:
∠BAC=2∠BCA,∠BCA+∠CAD=90∘andBC=BD. \angle BAC = 2\angle BCA, \quad \angle BCA + \angle CAD = 90^\circ \quad \text{and} \quad BC = BD.
Find ∠ADB\angle ADB.

Solution

Let ∠BCA=α\angle BCA = \alpha and XX be a point on ray CACA such that BX=BCBX = BC.
Since BX=BCBX = BC, we have that ∠BXC=α\angle BXC = \alpha, and using that ∠BAC=∠AXB+∠ABX\angle BAC = \angle AXB + \angle ABX, we obtain that ∠ABX=α\angle ABX = \alpha and hence AX=ABAX = AB.
We can observe that ∠XAD=∠DAB\angle XAD = \angle DAB, as ∠XAD=180∘−∠DAC=90∘+α\angle XAD = 180^\circ - \angle DAC = 90^\circ + \alpha, ∠DAB=90∘−α+2α=90∘+α\angle DAB = 90^\circ - \alpha + 2\alpha = 90^\circ + \alpha. Therefore, using SAS congruence we have △XAD≅△BAD\triangle XAD \cong \triangle BAD, and XD=DBXD = DB. This shows that △XDB\triangle XDB is equilateral, and ∠XDA=∠ADB=30∘\angle XDA = \angle ADB = 30^\circ.

Figure 1

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