Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 6.1 National olympiad Prove it Romania

Consider the isosceles right triangle $ABC$ , with $m(\widehat{BAC}) = 90^\circ$ . Take now the point $D$ so that $BD \perp BC$ and $AD = BC$ . Find the measure of the angle $\widehat{BAD}$ .

Solution

Case 1: $D$ and $A$ are on different sides of $BC$ (figure 1).
Denote $\{E\} = AC \cap DB$ . Then $m(\widehat{ABE}) = 45^\circ$ , therefore $[BA]$ is bisector and altitude in the triangle $BEC$ . So, $[AE] = [AC]$ . Construct $AM \perp BE$ , Then $[AM]$ is a midline in the triangle $EBC$ , hence $AM = \frac{1}{2}BC = \frac{1}{2}AD$ .
In the right triangle $MAD$ the leg $[AM]$ is half of the hypotenuse $[AD]$ , hence $m(\widehat{ADB}) = 30^\circ$ . It follows $m(\widehat{BAD}) = 180^\circ - m(\widehat{ADB}) - m(\widehat{ABD}) = 180^\circ - 30^\circ - 105^\circ = 15^\circ$ .

Case 2: $D$ and $A$ are on different sides of the line $BC$ (figure 2).
Denote $\{E\} = AC \cap DB$ . Then $m(\widehat{ABE}) = 45^\circ$ , hence $[BA]$ is bisector and altitude in the triangle $BEC$ . So, $[AE] = [AC]$ . Construct $AM \perp BE$ . Then $[AM]$ is a midline in the triangle $EBC$ , hence $AM = \frac{1}{2}BC = \frac{1}{2}AD$ .
In the right triangle $MAD$ the leg $[AM]$ is half of the hypotenuse $[AD]$ , hence $m(\widehat{ADB}) = 30^\circ$ . It follows $m(\widehat{BAD}) = 180^\circ - m(\widehat{ADB}) - m(\widehat{ABD}) = 180^\circ - 30^\circ - 45^\circ = 105^\circ$ .

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