Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that: AB/PQ = AD/PM. - Noon Academy

Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that: AB/PQ = AD/PM.

Class 10 | Exercise 15E | ICSE | Maths | Selina | Similarity (With Applications to Maps and Models)

Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that: AB/PQ = AD/PM.

Selina Solutions Concise Class 10 Maths Chapter 15 ex. 15(E) - 4

Solution:

According to the given question,

$\vartriangle ABC\text{ }\sim\text{ }\vartriangle PQR$

$AD\text{ }and\text{ }PM$

are the medians, so

$BD\text{ }=\text{ }DC\text{ }and\text{ }QM\text{ }=\text{ }MR$

So, we have

$AB/PQ\text{ }=\text{ }BC/QR$

[Corresponding sides of similar triangles are proportional.]

Then,

$AB/PQ\text{ }=\text{ }\left( BC/2 \right)/\text{ }\left( QR/2 \right)$

$=\text{ }BD/QM$

And,

$\angle ABC\text{ }=\angle PQR\text{ }i.e.\angle ABD\text{ }=\angle PQM$

So,

$\vartriangle ABD\text{ }\sim\text{ }\vartriangle PQM\text{ }by\text{ }SAS$

criterion for similarity.

Thus,

$AB/PQ\text{ }=\text{ }AD/PM$

i

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