Geometry (1)

Geometry Level 2

$\triangle ABC$ is inscribed to a circle with center $O$ . Let the ray $AO$ meet $BC$ at $D$ . Choose a point $K$ on the tangent line from point $A$ such that $\angle KBC=90^\circ$ .

Is $KD~\text{parallel to}~AC$ ?

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1 solution

Marta Reece
May 28, 2018

ADBK is a rectangle, since angle DBK is $90^\circ$ and so is angle KAD, the line AK being tangent to the circle and line AD going through the center of the circle.

This means that AK $=$ DB and the two are parallel to each other.

CD also equals DB, since OD is perpendicular to CB.

So AKDC is a parallelogram and AC is parallel to KD.

I think it would be more understandable if you said: "Choose a point $K$ on the line tangent to the circle at point $A$ such that $\angle KBC=90^\circ$ ."

Marta Reece - 3 years ago

Geometry (1)

1 solution

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