side ab and ac and median ad of a triangle abc r respectively proportional to sides pq and pr and median pm of another triangle pqr prove that triangle abc similar to pqr explain step by step it plz
GIVEN - AB/PQ , AC/PR , AD/PM
CONSTRUCTION - Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L.
PROVE - ΔABC ∼ ΔPQR
We know that medians divide opposite sides.
Therefore, BD = DC and QM = MR
Also, AD = DE (By construction)
And, PM = ML (By construction)
In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.
Therefore, quadrilateral ABEC is a parallelogram.
∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)
Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR
It was given that
∴ ΔABE ∼ ΔPQL (By SSS similarity criterion)
We know that corresponding angles of similar triangles are equal.
∴ ∠BAE = ∠QPL … (1)
Similarly, it can be proved that ΔAEC ∼ ΔPLR and
∠CAE = ∠RPL … (2)
Adding equation (1) and (2), we obtain
∠BAE + ∠CAE = ∠QPL + ∠RPL
⇒ ∠CAB = ∠RPQ … (3)
In ΔABC and ΔPQR,
(Given)
∠CAB = ∠RPQ [Using equation (3)]
∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)
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