Up until now I’ve been using enlargements as a shortcut for similar triangles, but it was still kind of wordy having to state the centre and scale factor to justify why two lines might be parallel. I just found out that this is known as the intercept theorem (and its converse) – now I can just refer to this theorem every time! My life just got so much better.

**ABC is a triangle with AB = 2BC. M is the midpoint of AB and P is on the segment AC such that AC = 3AP. Then PMC = 90.**

**Proof 1**

Choose D on AB produced beyond B such that MB = BD.

Then as BM = BC = BD, B is the circumcentre of triangle MCD. Since B lies on MD, MD is a diameter of MCD’s circumcircle so ∠MCD = 90.

But CD is parallel to PM by the converse of the intercept theorem as AC = 3AP and AD = 3AM.

Hence ∠PMC = 90 by alternate angles.

**Proof 2**

Choose D on BC produced beyond C such that BC = CD. Let Q be the midpoint of PC, so AP = PQ = QC.

Then by the converse of the intercept theorem, BQ is parallel to MP and MC is parallel to AD.

But now consider triangle ABD. AC is a median and Q is the point on the median such that AQ = 2QC. Thus Q is the centroid of triangle ABD.

Also, BD = 2BC = AB so triangle ABD is isosceles. Thus the median from B is perpendicular to AD, so BQ is perpendicular to AD.

Using our parallelisms, this implies MP and MC are perpendicular as well.

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