# How large is angle CKA?

Triangle ABC has angles A=72°, B=60°, C=48°.

A circle is inscribed in a triangle ABC. The circle touches AB and BC at points E and F, respectively. Angle bisector A intersects line EF at point K.

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• Duke
Lv 7

The answer is angle(CKA) = 90°.

Let I is the incenter of ∆ABC (intersection point of angle bisectors of A and B). Since both tangents from B to the incircle are equal, ∆BEF is equilateral hence

angle(AEF) = angle(AEK) = 120°, angle(BAK) = 36°, angle(AKE) = 24°

Consider ∆AEK and ∆AIC:

angle(EAK) = angle(IAC) = 36°, angle(AKE) = angle(ACI) = 24°,

angle(AEK) = angle(AIC) = 120°, i.e. they are similar, the coefficient of similarity is

| AE | : | AI | = cos 36° (∆AEI is right)

Then also | AK | : | AC | = cos 36°, but the latter is possible only if ∆AKC is right.

By the way K is outside ∆ABC because the line AK intersects BC in a point let's say A' such that ∆ABA' has angles 36°, 60°, 84° and IF as inradius is perpendicular to BC.

• 1 decade ago

Do you have a picture for the problem? It's hard for me to know whether K is inside or outside of the circle.

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I just figured out that K must be outside the circle.

Duke got it done before me.

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Here is my approach:

Let O be the incenter, and T be the intersection point of the angle bisector and BC.

First, we can prove that

∆COT ~ ∆KFT because angle OCT = 24 degrees, and angle FKT = 180-36-120 = 24 degrees, and angle CTO and angle KTF are vertical angles.

Therefore,

OT:FT = CO:KF

Since angle CTK and angle OTF are vertical angles, they are congruent.

So, we have

∆CTK ~ ∆OTF (by SAS ~ theorem)

angle CKT = angle OFT = 90 degrees

angle CKA = angle CKT = 90 degrees