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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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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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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