If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, proves that \(\frac{AB}{PQ}=\frac{AD}{PM}\)
Solution and Explanation
Given: AD and PM are medians of triangles ABC and PQR
ΔABC ~ ΔPQR
To Prove: \(\frac{AB}{PQ}=\frac{AD}{PM}\)
Proof: It is given that ∆ABC ∼ ∆PQR
We know that the corresponding sides of similar triangles are in proportion.
∴\(\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}\) … (1)
Also, \(\angle\)A = \(\angle\)P, \(\angle\)B = \(\angle\)Q, \(\angle\)C = \(\angle\)R … (2)
Since AD and PM are medians, they will divide their opposite sides.
∴BD=\(\frac{BC}{2}\) and QM=\(\frac{QR}{2}\) … (3)
From equations (1) and (3), we obtain
\(\frac{AB}{PQ}=\frac{BD}{QM}\) … (4)
In ∆ABD and ∆PQM,
\(\angle\)B = \(\angle\)Q [Using equation (2)]
\(\frac{AB}{PQ}=\frac{BD}{QM}\)[Using equation (4)]
∴ ∆ABD ∼ ∆PQM (By SAS similarity criterion)
⇒ \(\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}\)
\(\therefore\frac{AB}{PQ}=\frac{AD}{PM}\)
Hence Proved
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