The 80-80-20 Triangle Problem, Solution #10

Let ABC be an isosceles triangle (AB = AC) with ∠BAC = 20°. Point D is on side AC such that ∠CBD = 50°. Point E is on side AB such that ∠BCE = 60°. Find the measure of ∠CED.

Solution

The proof is due to Luke Rapley.

With radius CE and center C sweep an arc to intersect BC at U and AC at V.

ΔCEU is isosceles (CE = CU) and, ∠BCE being 60°, is also equilateral. In particular, EU = CE and ∠CEU = 60°. It follows that ∠BEU = 20°. Another preliminary result, viz., AE = CE, follows from the observation that ΔACE having angles at A and C both equal to 20° is isosceles.

We are going to compare triangles BEU and VAE:

AE = EU, since both equal CE,
AV = BE, because AV = AC - CV = AB - AE = BE,
∠EAV = ∠BEU since both equal 20°.

We conclude that, by SAS, the two triangles are equal. From here we get EV = BU.

On the other hand, since ΔBCD is isosceles, BC = CD and consequently,

BU = CU - BC = CV - CD = DV.

Therefore, VD = EV and triangle DEV is isosceles. Now,

∠DVE = 180° - ∠AVE = 180° - ∠EBU = ∠ABC = 80°.

So ∠DEV = (180° - 80°)/2 = 50°. Counting angles around E, we see that ∠CED = 30°.

The 80-80-20 Triangle Problem

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