Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ABC ~ PQR.
Step-by-Step Solution
Key Concept: If two sides of a triangle are proportional to two sides of another triangle and the median to the third side is also proportional, then the third sides are proportional as well. Hence all three corresponding sides are in the same ratio, which gives similarity by the SSS criterion.
1. Given proportionalities
$$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}=k \quad (k>0)$$
Hence
$$AB=k\,PQ,\qquad AC=k\,PR,\qquad AD=k\,PM.$$
2. Use Apollonius theorem for triangle $ABC$ (median $AD$ to side $BC$):
$$AB^{2}+AC^{2}=2\bigl(AD^{2}+\left(\frac{BC}{2}\right)^{2}\bigr)\;\;\Rightarrow\;\;AB^{2}+AC^{2}=2AD^{2}+\frac{BC^{2}}{2} \tag{1}$$
Similarly, for triangle $PQR$ (median $PM$ to side $QR$):
$$PQ^{2}+PR^{2}=2PM^{2}+\frac{QR^{2}}{2} \tag{2}$$
3. Substitute the proportionalities from step 1 into (1):
\[
(kPQ)^{2}+(kPR)^{2}=2(kPM)^{2}+\frac{BC^{2}}{2}
\]
Simplify by dividing by $k^{2}$:
\[
PQ^{2}+PR^{2}=2PM^{2}+\frac{BC^{2}}{2k^{2}} \tag{3}
\]
4. Compare (3) with (2). Since the left‑hand sides of (2) and (3) are identical, the right‑hand sides must be equal:
\[
2PM^{2}+\frac{QR^{2}}{2}=2PM^{2}+\frac{BC^{2}}{2k^{2}}
\]
Hence
\[
\frac{QR^{2}}{2}=\frac{BC^{2}}{2k^{2}}\;\Rightarrow\; QR = k\,BC.
\]
Thus the third sides are also in the same ratio $k$:
$$\frac{BC}{QR}=k.$$
5. All three corresponding sides are proportional:
$$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}=k.$$
By the SSS similarity criterion, the two triangles are similar:
$$\boxed{\Delta ABC \sim \Delta PQR}.$$
6. Conclusion – The given proportionalities of two sides and the median force the third side to be proportional, establishing the similarity of the two triangles.
Correct Answer: Δ ABC ~ Δ PQR