In the given figure, ΔABC ∼ ΔPQR, AM and PN are altitudes, whereas AX and PY are medians. Prove that $\dfrac{AM}{PN} = \dfrac{AX}{PY}$.

Since ΔABC ∼ ΔPQR

So, their respective sides will be in proportion

$\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{BC}{QR}$

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R

In ΔABM and ΔPQN,

∠ABM = ∠PQN [Since, ABC and PQR are similar]

∠AMB = ∠PNQ = 90° [Given ]

∴ ΔΑΒΜ ∼ ΔPQN by AA similarity

$\dfrac{AM}{PN} = \dfrac{AB}{PQ}$ …..(1)

Since, AX and PY are medians so they will divide their opposite sides.

BX = $\dfrac{BC}{2}$ and QY = $\dfrac{QR}{2}$

Therefore, we have:

$\dfrac{AB}{PQ} = \dfrac{BX}{QY}$

∠ABC = ∠PQR

So, we had observed that two respective sides are in same proportion in both triangles and also angle included between them is respectively equal.

Hence, ∆ABX ∼ ∆PQY (by SAS similarity rule).So,

$\dfrac{AM}{PQ} = \dfrac{AX}{PY}$ …..(2)

From (1) and (2),

$\dfrac{AM}{PN} = \dfrac{AX}{PY}$

Hence, proved that $\dfrac{AM}{PN} = \dfrac{AX}{PY}$