Find the x in the problem

Geometry Level pending


The answer is 20.

Aly Ahmed by Aly Ahmed , 34, Egypt

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1 solution

Let equilateral Δ A E C \Delta AEC be escribed upon A C AC . Δ B I C ∼ Δ B C E ⇒ B C I C = B E C E = C D A C \Delta BIC \sim \Delta BCE \Rightarrow \dfrac{BC}{IC}=\dfrac{BE}{CE} =\dfrac{CD}{AC} [Ptolemy's in equil. Δ \Delta ]

Also, ∠ A I C = ∠ C B D = 15 0 ∘ ⇒ S S A . ∼ Δ I C A ∼ Δ D B C ⇒ x = 2 0 ∘ Q . E . D \angle AIC= \angle CBD =150^{\circ} \stackrel{SSA.\sim}\Rightarrow \Delta ICA \sim \Delta DBC \Rightarrow x=20^{\circ} \boxed{Q.E.D}

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