Here is an interesting result I discovered that can be proven using homotheties, or enlargements.

**Let the incircle ω of triangle ABC touch BC at R. A line s is drawn parallel to (and distinct from) BC such that it is tangent to ω. Let s cut AB and AC at P and Q respectively. Then BQ and CP meet on the diameter of ω that passes through R.**

Proof

Let s touch ω at X. This point is the other end of the diameter through R since s is parallel to BC. We wish to show that RX, BQ and CP are concurrent.

Since s it parallel to BC, triangle ABC is an enlargement of triangle APQ, centre A; let this enlargement, sending APQ to ABC, have scale factor k.

Extend AX to meet BC at Z.

You may recognise this famous configuration. ω is the excircle of APQ opposite A (the A-excircle), so the enlargement centre A and scale factor k must carry the point of contact between PQ and the A-excircle of APQ to the point of contact between BC and the A-excircle of ABC. Since X is carried to Z by this homothety, Z is the point of contact between BC and the A-excircle of ABC.

We claim BZ = CR; this is a well-known fact. To prove this, consider this separate diagram:

Let AE = AF = x (equal tangents); similarly, let BD = BF = y and let CD = CE = z.

The perimeter of ABC is 2(x + y + z).

Since BZ = BX and CX = CY by equal tangents, AZ = AB + BX and AY = AC + CX.

But AY = AZ, since they are tangents to the A-excircle and hence equal; and as AY + AZ = AB + BX + AC + CX = 2(x + y + z), we must have AY = AZ = x + y + z.

So BX = BZ = AZ – AB = x + y + z – (x + y) = z = CD

So BX = CD.

Returning to our original diagram,

We have BZ = CR. Therefore, as BZ = kPX, CR = kPX.

Also, QX = PQ – PX, so BR = BC – CR = kPQ – kPX = k(PQ – PX) = kQX.

Also, since PX is parallel to CR, there exists an inverse homothety that sends PX to CR. The centre of this homothety (scale factor -k) must be at the intersection of RX and CP – therefore this point divides line segments RX and CP in the ratio 1 : k.

Similarly, there exists an inverse homothety, scale factor -k, that sends QX to BR, whose centre, located at the intersection of RX and BQ, divides the line segments RX and BQ in the ratio 1 : k. But there can only be one point on RX which divides RX in the ratio 1 : k, so the centres of the two homotheties are the same point and hence these 3 lines must be concurrent at this point.

### Like this:

Like Loading...