An isosceles triangle $ABC$ has $AB=AC$. Angle $A$ measures $20^\circ$. On $AC$, point $E$ is such that $AE=BC$. The task is to find the measure of angle $BEC$ without using trigonometry. How can one go about this?
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1Point E is on AC. EAC will be 0. – Satvik Mashkaria Apr 13 '14 at 17:45
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I can't see how knowing $;BC=AE;$ can be helpful without trigonometry in this case... – DonAntonio Apr 13 '14 at 17:47
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Investigate Euclid's Elements Book 1 Proposition 5 for an idea of how to proceed. – JEM Apr 13 '14 at 17:49
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Its simple angle chasing... – Sawarnik Apr 13 '14 at 17:54
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Construct an isosceles triangle AEQ with ∠AQE = 20°. Since. AE = BC, the latter is equal to ΔABC. In particular, AQ = AB. Also,
∠BAQ = ∠EAQ - ∠EAB = 80° - 20° = 60°.
Which makes ΔABQ equilateral. In particular, BQ = EQ = AQ.
In ΔBQE, EQ = BQ and ∠BQE = 60° - 20° = 40°. Thus, ∠BEQ = (180° - 40°) / 2 = 70° so that ∠AEC = 80° + 70° = 150°.and from that angle BEC=30.
Taha Akbari
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