Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that \(\triangle ABC \sim \triangle PQR\).
Produce AD to a point E such that AD = DE and produce PM to a point N such that PM = MN. Join EC and NR.
Given: In \(\triangle ABC\) and \(\triangle PQR\), AD and PM are medians and
\[\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}.\]
To prove: \(\triangle ABC \sim \triangle PQR\).
Produce AD to a point E such that \(AD = DE\). Produce PM to a point N such that \(PM = MN\). Join BE, CE, QN and RN.
In \(\triangle ABD\) and \(\triangle CDE\):
\(AD = DE\) (construction), \(BD = DC\) (AD is a median), and \(\angle ADB = \angle CDE\) (vertically opposite).
So \(\triangle ABD \cong \triangle CDE\) (SAS), hence \(AB = CE\) and D is the midpoint of both BC and AE. Thus, diagonals BC and AE bisect each other, so quadrilateral ABEC is a parallelogram. Therefore,
\[AB \parallel CE, \quad BC \parallel AE.\]
Similarly, in \(\triangle PQM\) and \(\triangle MNR\):
\(PM = MN\) (construction), \(QM = MR\) (PM is a median), and \(\angle PMQ = \angle NMR\) (vertically opposite),
so \(\triangle PQM \cong \triangle NRM\). Hence \(PQ = RN\) and M is the midpoint of both QR and PN, so PQNR is a parallelogram and
\[PQ \parallel RN, \quad QR \parallel PN.\]
From the given proportionality,
\[\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}.\]
Using \(AB = CE\) and \(PQ = RN\) (from the congruent triangles above), we get
\[\frac{CE}{RN} = \frac{AC}{PR} = \frac{AD}{PM}.\]
Also, \(AE = 2AD\) and \(PN = 2PM\), so
\[\frac{AE}{PN} = \frac{2AD}{2PM} = \frac{AD}{PM}.\]
Hence,
\[\frac{CE}{RN} = \frac{AC}{PR} = \frac{AE}{PN}.\]
Thus, all three pairs of corresponding sides in \(\triangle ACE\) and \(\triangle PRN\) are proportional, so
\[\triangle ACE \sim \triangle PRN \quad (\text{SSS similarity}).\]
From \(\triangle ACE \sim \triangle PRN\), we have
\[\angle CAE = \angle RPN.\]
Since ABEC and PQNR are parallelograms, \(AB \parallel CE\) and \(QP \parallel RN\). Therefore, angle between BA and AE equals angle between CE and EN (or RN and NP):
\[\angle BAE = \angle QPN.\]
Now, angle at A of \(\triangle ABC\) can be split as
\[\angle BAC = \angle CAE + \angle BAE,\]
and angle at P of \(\triangle PQR\) as
\[\angle QPR = \angle RPN + \angle QPN.\]
Using \(\angle CAE = \angle RPN\) and \(\angle BAE = \angle QPN\), we get
\[\angle BAC = \angle QPR,\]
so
\[\angle A = \angle P.\]
We already know
\[\frac{AB}{PQ} = \frac{AC}{PR}\]
and we have just proved \(\angle A = \angle P\). Thus, by the SAS similarity criterion,
\[\triangle ABC \sim \triangle PQR.\]
Hence proved.