If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that AB/PQ = AD/PM
Solution:
We know if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
This is referred as SAS criterion for two triangles.
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,
∠ABD = ∠PQM (from 1)
AB/PQ = BD/QM (from 2)
⇒ ΔABD ∼ ΔPQM (SAS criterion)
⇒ AB/PQ = BD/QM = AD/PM (corresponding sides)
⇒ AB/PQ = AD/PM
Hence proved.
☛ Check: NCERT Solutions for Class 10 Maths Chapter 6
Video Solution:
If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that AB/PQ = AD/PM
NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 Question 16
Summary:
If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, it is proved that AB/PQ = AD/PM.
☛ Related Questions:
- In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ΔABC ~ ΔAMP (ii) CA/PA = BC/MP.
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