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Triangles
EXERCISE 6.3
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
Sides AB and BC and median AD of a triangle ABC are respectively propor- tional to sides PQ and QR and median PM of PQR (see Fig. 6.41). Show that ABC ~ PQR.
Step-by-Step Solution
Key Concept: Use the SSS similarity criterion. From the given proportionalities of two sides and the corresponding medians, apply Apollonius’ theorem (which relates a side, the median to the opposite side and the other two sides) to obtain the proportionality of the third side. Once all three corresponding sides are shown to be in the same ratio, the triangles are similar by SSS.
1. Given proportionalities\
\[\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}=k\] \(k>0\) is a constant.
2. Express the median using Apollonius’ theorem\
For ΔABC, the median AD to side BC satisfies\
\[AB^{2}+AC^{2}=2\bigl(AD^{2}+BD^{2}\bigr)\]\
where \(BD=DC=\dfrac{BC}{2}\).
For ΔPQR, the median PM to side QR satisfies\
\[PQ^{2}+PR^{2}=2\bigl(PM^{2}+QM^{2}\bigr)\]\
where \(QM=MR=\dfrac{QR}{2}\).
3. Replace the sides of ΔPQR by the corresponding sides of ΔABC using the ratio \(k\)\
\[PQ = \frac{AB}{k},\qquad QR = \frac{BC}{k},\qquad PM = \frac{AD}{k},\qquad QM = \frac{BC}{2k}.\]
4. Substitute these expressions in the Apollonius relation for ΔPQR\
\[\left(\frac{AB}{k}\right)^{2}+PR^{2}=2\left[\left(\frac{AD}{k}\right)^{2}+\left(\frac{BC}{2k}\right)^{2}\right].\]
Multiply by \(k^{2}\):\
\[AB^{2}+k^{2}PR^{2}=2\bigl(AD^{2}+\frac{BC^{2}}{4}\bigr).\]
5. Use the Apollonius relation for ΔABC\
\[AB^{2}+AC^{2}=2\bigl(AD^{2}+\frac{BC^{2}}{4}\bigr).\]
Comparing the right‑hand sides of the two equations we obtain\
\[AB^{2}+k^{2}PR^{2}=AB^{2}+AC^{2}\]\
Hence\
\[k^{2}PR^{2}=AC^{2}\]\
or\
\[\frac{AC}{PR}=k.\]
6. Now all three corresponding sides are in the same ratio\
\[\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=k.\]
Therefore, by the SSS similarity criterion,\
\[\Delta ABC \sim \Delta PQR.\]
7. Conclusion\
The given proportionality of two sides and the corresponding medians forces the third side to be in the same ratio, establishing the similarity of the two triangles.
\[\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}=k\] \(k>0\) is a constant.
2. Express the median using Apollonius’ theorem\
For ΔABC, the median AD to side BC satisfies\
\[AB^{2}+AC^{2}=2\bigl(AD^{2}+BD^{2}\bigr)\]\
where \(BD=DC=\dfrac{BC}{2}\).
For ΔPQR, the median PM to side QR satisfies\
\[PQ^{2}+PR^{2}=2\bigl(PM^{2}+QM^{2}\bigr)\]\
where \(QM=MR=\dfrac{QR}{2}\).
3. Replace the sides of ΔPQR by the corresponding sides of ΔABC using the ratio \(k\)\
\[PQ = \frac{AB}{k},\qquad QR = \frac{BC}{k},\qquad PM = \frac{AD}{k},\qquad QM = \frac{BC}{2k}.\]
4. Substitute these expressions in the Apollonius relation for ΔPQR\
\[\left(\frac{AB}{k}\right)^{2}+PR^{2}=2\left[\left(\frac{AD}{k}\right)^{2}+\left(\frac{BC}{2k}\right)^{2}\right].\]
Multiply by \(k^{2}\):\
\[AB^{2}+k^{2}PR^{2}=2\bigl(AD^{2}+\frac{BC^{2}}{4}\bigr).\]
5. Use the Apollonius relation for ΔABC\
\[AB^{2}+AC^{2}=2\bigl(AD^{2}+\frac{BC^{2}}{4}\bigr).\]
Comparing the right‑hand sides of the two equations we obtain\
\[AB^{2}+k^{2}PR^{2}=AB^{2}+AC^{2}\]\
Hence\
\[k^{2}PR^{2}=AC^{2}\]\
or\
\[\frac{AC}{PR}=k.\]
6. Now all three corresponding sides are in the same ratio\
\[\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=k.\]
Therefore, by the SSS similarity criterion,\
\[\Delta ABC \sim \Delta PQR.\]
7. Conclusion\
The given proportionality of two sides and the corresponding medians forces the third side to be in the same ratio, establishing the similarity of the two triangles.
Correct Answer: ΔABC is similar to ΔPQR (ΔABC ~ ΔPQR).
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