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Question
In the given figure, triangle ABC is similar to triangle PQR. AM and PN are altitudes whereas AX and PY are medians.
prove that
`("AM")/("PN")=("AX")/("PY")`
Solution
Since ΔABC ~ ΔPQR
So, their respective sides will be in proportion
Or `("AB")/("PQ")=("AC")/("PR")=("BC")/("QR")`
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R
In ∆ABM and ∆PQN,
∠ABM = ∠PQN (Since, ABC and PQR are similar)
∠AMB = ∠PNQ = 90°
∆ABM ~ ∆PQN (AA similarity)
∴`("AM")/("PN")=("AB")/("PQ")`..........................................(1)
Since, AX and PY are medians so they will divide their opposite sides.
Or , `"BX"=("BC")/2 ` and `"QY"=("QR")/(2) `
Therefore, we have:
`("AB")/("PQ")=("BX")/("QY")`
∠B = ∠Q
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. `("AM")/("PQ")=("AX")/("PY")` ......................(2)
From (1) and (2),
`("AM")/("PN") = ("AX")/("PY")`