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Question: In the given figures sides \(AB\) and \(BC\) and median \(AD\) of a \(\Delta ABC\) are respectively ...

In the given figures sides ABAB and BCBC and median ADAD of a ΔABC\Delta ABC are respectively proportional to sides PQ,QRPQ,QR and median PMPM of ΔPQR\Delta PQR . show that triangle ΔABCΔPQR\Delta ABC \sim \Delta PQR.

Explanation

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,
ABPQ=BCQR=ADPM(1)\dfrac{{AB}}{{PQ}} = \dfrac{{BC}}{{QR}} = \dfrac{{AD}}{{PM}} - - - - - \left( 1 \right)
In ΔABC\Delta ABC, since ADAD is the median,
BD=CD=12BCBD = CD = \dfrac{1}{2}BC
Or BC=2BD(2)BC = 2BD - - - - - \left( 2 \right)
Similarly, PMPM is the median,
QM=RM=12QRQM = RM = \dfrac{1}{2}QR
Or QR=2QM(3)QR = 2QM - - - - - \left( 3 \right)
Substituting the value of BC,QRBC,QR in equation (1), we get
ABPQ=2BD2QM=ADPM\dfrac{{AB}}{{PQ}} = \dfrac{{2BD}}{{2QM}} = \dfrac{{AD}}{{PM}}
ABPQ=BDQM=ADPM(4)\dfrac{{AB}}{{PQ}} = \dfrac{{BD}}{{QM}} = \dfrac{{AD}}{{PM}} - - - - - \left( 4 \right)
Since all three sides are proportional.
Therefore, by SSS similarity rule,
ΔABDΔPQM\Delta ABD \sim \Delta PQM
Hence,B=Q(5)\angle B = \angle Q - - - - - - \left( 5 \right),
corresponding angles of similar triangles are equal.
In ΔABC&ΔPQR\Delta ABC\& \Delta PQR
Using (5), we get
B=Q\angle B = \angle Q
Given, ABPQ=BCQR\dfrac{{AB}}{{PQ}} = \dfrac{{BC}}{{QR}}
Hence by SAS Similarity of triangle.
ΔABCΔPQR\Delta ABC \sim \Delta PQR

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.