Verified answer

★ Firstly try understanding the diagram.

Here it is given that,

★ AD and PM are medians of triangles △ABC and △PQR respectively.

• And, as ∆ ABC ~ ∆ PQR

So, the following things can be said:

[tex] \sf \implies \left( \dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR} \right) \: \: (eq ^{n} 1)[/tex]

Also,

[tex] \sf \implies \angle A = \angle P, \: \angle B = \angle Q \: and \: \angle C = \angle R [/tex]

Now,

In equation 1, BC can be written as 2BD. The reason for the same is, AD⊥BC. Similarly, QR can be written as 2QM.

[tex] \sf \implies \dfrac{AB}{PQ} = \dfrac{2BD}{2QM} = \dfrac{AC}{PR} [/tex]

[tex] \sf \implies \dfrac{AB}{PQ} = \dfrac{BD}{QM} [/tex]

Now,

[tex]\sf \implies \angle B = \angle Q [/tex]

So, by SAS!

• ∆ ABD ~ ∆ PQM

Therefore,

[tex] \implies \boxed{ \sf \dfrac{AB }{PQ } = \dfrac{ AD }{PM } }[/tex]

Hence Proved

QuestioN:

• If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ~ ∆PQR, prove that

[tex] \dfrac{AB }{PQ } = \dfrac{ AD }{PM } [/tex]

GiveN:

ΔABC ∼ ΔPQR

★⇒ ∠ABC = ∠PQR (corresponding angles) ---------(1)

★⇒ AB/PQ = BC/QR (corresponding sides)

★⇒ AB/PQ = (BC/2) / (QR/2)

★⇒ AB/PQ = BD/QM (D and M are mid-points of BC and QR) ------------ (2)

In ΔABD and ΔPQM,

[tex]\implies∠ABD = ∠PQM (from 1)[/tex]

[tex]\implies \: AB/PQ = BD/QM (from 2)[/tex]

[tex] \\\implies ΔABD ∼ ΔPQM \\ (SAS criterion)[/tex]

[tex]\implies AB/PQ = BD/QM = AD/PM \\ (corresponding \: sides)[/tex]

[tex]\implies \: AB/PQ = AD/PM[/tex]

Hence proved.