CA UD.UU.
19 Sides AB and AC and median AD of a
triangle ABC are respectively
proportional to sid:s PQ and PR and
median PM of another triangle PQR.
Show that A ABC-A POR.
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Answer: check answer in the given attachment
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Given two triangles. ΔABC and ΔPQR in which AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR
AB/PQ = BC/QR = AD/PM
To Prove: ΔABC ~ ΔPQR
Proof: AB/PQ = BC/QR = AD/PM
AB/PQ = BC/QR = AD/PM (D is the mid-point of BC. M is the mid point of QR)
ΔABD ~ ΔPQM [SSS similarity criterion]
Therefore, ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]
∠ABC = ∠PQR
In ΔABC and ΔPQR
AB/PQ = BC/QR ———(i)
∠ABC = ∠PQR ——-(ii)
From above equation (i) and (ii), we get
ΔABC ~ ΔPQR [By SAS similarity criterion]
Hence Proved
HOPE IT HELPS U :)
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