Library / /10 of 69

, 2011

Geometry Difficulty 4.6 AIME Prove it South Africa

In a convex quadrilateral $ABCD$ the sides $AB$ , $BC$ , $CD$ are equal and $AC = BD = AD$ . Find the angles of $ABCD$ .

Solution

Notice that $\triangle ABC \equiv \triangle BCD$ (s, s, s) and $\triangle ABD \equiv \triangle DCA$ (s, s, s).
Let $\alpha = \angle CBD = \angle BCA = \angle CAB = \angle CDB$
and $\beta = \angle ACD = \angle ADC = \angle BAD = \angle ABD$ .

It can then be seen that $3\alpha + \beta = 180^\circ$ from triangle $ABC$ and $4\beta + 2\alpha = 360^\circ$ from quadrilateral $ABCD$ . Solving these two simultaneous equations gives $\alpha = 36^\circ$ and $\beta = 72^\circ$ . Hence $\angle A = 72^\circ$ , $\angle B = 108^\circ$ , $\angle C = 108^\circ$ , $\angle D = 72^\circ$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.