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Question
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR show that ΔABC ~ ΔPQR.
Solution
We have MBC and PQR in which AD and PM are medians corresponding to sides BC and QR respectively such that
`("AB")/("PQ") = ("BC")/("QR") = ("AD")/("PM")`
`("AB")/("PQ") = (1/2 "BC")/(1/2 "QR") = ("AD")/("PM")`
`("AB")/("PQ") = ("BD")/("QM") = ("AD")/("PM")`
Using SSS similarity, we have
ΔABD ~ ΔPQM
Their corresponding angles are equal.
∠ABD = ∠PQM
∠ABC = ∠PQR
Now, in MBC and ΔPQR,
`("AB")/("PQ") = ("BC")/("QR")` ...[1]
Also, ∠ABC = ∠PQR ...[2]
From [1] and [2]
ΔABC ~ ΔPQR ...[SAS similarity]
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