Question
Question: In the given figures sides \(AB\) and \(BC\) and median \(AD\) of a \(\Delta ABC\) are respectively ...
In the given figures sides AB and BC and median AD of a ΔABC are respectively proportional to sides PQ,QR and median PM of ΔPQR . show that triangle ΔABC∼ΔPQR.
Solution
Hint: In order to solve this question, we have to apply similarity rules of triangles and in which side and angles helps us to show the similarities of these triangles.
Complete step-by-step answer:
According to given question,
PQAB=QRBC=PMAD−−−−−(1)
In ΔABC, since AD is the median,
BD=CD=21BC
Or BC=2BD−−−−−(2)
Similarly, PM is the median,
QM=RM=21QR
Or QR=2QM−−−−−(3)
Substituting the value of BC,QR in equation (1), we get
PQAB=2QM2BD=PMAD
PQAB=QMBD=PMAD−−−−−(4)
Since all three sides are proportional.
Therefore, by SSS similarity rule,
ΔABD∼ΔPQM
Hence,∠B=∠Q−−−−−−(5),
corresponding angles of similar triangles are equal.
In ΔABC&ΔPQR
Using (5), we get
∠B=∠Q
Given, PQAB=QRBC
Hence by SAS Similarity of triangle.
ΔABC∼ΔPQR
Note: Whenever we face these types of questions the key concept is that we have to take small triangles and by similarity rules show them similar by which we get two sides or one sides and one angle equality and we will easily get our desired answer.