Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see the given figure). Show that ΔABC ∼ ΔPQR.

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#### Solution

Median divides the opposite side.

`:. BD = (BC)/2 and QM = (QR)/2`

Given that

`(AB)/(PQ) = (BC)/(QR)=(AD)/(PM)`

`=>(AB)/(PQ) = (1/2(BC))/(1/2QR) = (AD)/(PM)`

`=> (AB)/(PQ) = (BD)/(QM) = (AD)/(PM)`

In ΔABD and ΔPQM,

`(AB)/(PQ) = (BD)/(QM)= (AD)/(PM) ` (Proved above)

∴ ΔABD ∼ ΔPQM (By SSS similarity criterion)

⇒ ∠ABD = ∠PQM (Corresponding angles of similar triangles)

In ΔABC and ΔPQR,

∠ABD = ∠PQM (Proved above)

`(AB)/(PQ) = (BC)/(QR)`

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

Concept: Criteria for Similarity of Triangles

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