Problem 5

Given three distinct points

A,B,C

on a circle

K

, construct a point

D

K

such that a circle can be inscribed in the convex quadrilateral

ABCD

Step 5 of 5: Justify that the construction is possible

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In plain words

A cyclic quadrilateral has an incircle exactly when the sums of opposite sides agree; continuity guarantees a point $D$ where this happens.

AB+CD=BC+AD

Detailed analysis

For $D$ moving on the chosen arc of $K$ , the continuous function $AB+CD-(BC+AD)$ is negative when $D$ is near $C$ and positive when $D$ is near $A$ . Hence it vanishes for some $D$ , and the standard tangent-length criterion for a convex quadrilateral shows that this is exactly the condition for an incircle. The preceding construction locates that $D$ via its incenter.