Question

If AD and PS are medians of ∆ABC and ∆PQR respectively where ∆ABC ~ ∆PQR, Prove that AB/ PQ = AD/ PS .

Answer 1

GIVEN: In ∆ABC & ∆PQR, D & S are the mid points of the sides BC & QR .
ΔABC ~ ΔPQR

AB/PQ = BC/QR = AC/PR ………...(i)
[corresponding sides of  two similar triangles are in proportional]

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …...…(ii)
[corresponding angles of two similar triangles are in equal]

BD = CD = 1/2 BC  and QS = SR = 1/2QR…….…...(iii)
[D is the midpoint of BC and S is the midpoint of QR]

AB/PQ = BC/QR   [From eq (i)]
AB/PQ = 2BD/2QS
[From eq iii ]

AB/PQ = BD/QS……………(iv)

In ΔABD and ΔPQS,
∠B = ∠Q         [From eq (ii)]
AB/PQ = BD/QS [From  eq (iv)]
ΔABD ~ ΔPQS (By SAS similarity criterion)
AB/PQ = AD/PS

[corresponding sides of similar triangles are proportional]

Hence, proved.

HOPE THIS WILL HELP YOU...

Answer 2

Answer:

Step-by-step explanation:

GIVEN: In ∆ABC & ∆PQR, D & S are the mid points of the sides BC & QR .

ΔABC ~ ΔPQR

AB/PQ = BC/QR = AC/PR ………...(i)

[corresponding sides of  two similar triangles are in proportional]

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …...…(ii)

[corresponding angles of two similar triangles are in equal]

BD = CD = 1/2 BC  and QS = SR = 1/2QR…….…...(iii)

[D is the midpoint of BC and S is the midpoint of QR]

AB/PQ = BC/QR   [From eq (i)]

AB/PQ = 2BD/2QS

[From eq iii ]

AB/PQ = BD/QS……………(iv)

In ΔABD and ΔPQS,

∠B = ∠Q         [From eq (ii)]

AB/PQ = BD/QS [From  eq (iv)]

ΔABD ~ ΔPQS (By SAS similarity criterion)

AB/PQ = AD/PS

[corresponding sides of similar triangles are proportional]

Hence, proved.