CBSE2026Class Class 10 · Mathematics5 Marks · Long✅ Verified
(a) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
OR
(b) AD and PS are respectively, the medians of ΔABC and ΔPQR. If ΔABC ~ ΔPQR, then prove that (i) ΔADC ~ ΔPSR (ii) AD/PS = BC/QR
✅ Answer & Solution
**(a)** (Basic Proportionality Theorem) **Given:** ΔABC with DE || BC, D on AB, E on AC. **To Prove:** $\frac{AD}{DB}=\frac{AE}{EC}$. **Construction:** Join BE, CD; draw $DM\perp AC$, $EN\perp AB$.
$$\frac{ar(\triangle ADE)}{ar(\triangle DBE)}=\frac{AD}{DB},\quad \frac{ar(\triangle ADE)}{ar(\triangle DEC)}=\frac{AE}{EC}$$
Since ΔDBE and ΔDEC have equal areas (same base DE, between same parallels), $\frac{AD}{DB}=\frac{AE}{EC}$. Hence proved.
**(b)** Since ΔABC ~ ΔPQR: $\angle B=\angle Q$ and $\frac{AB}{PQ}=\frac{BC}{QR}$. Since AD, PS are medians, $BD=\frac{BC}{2}$, $QS=\frac{QR}{2}$, so $\frac{BD}{QS}=\frac{BC}{QR}=\frac{AB}{PQ}$.
(i) In ΔABD and ΔPQS: $\angle B=\angle Q$, $\frac{AB}{PQ}=\frac{BD}{QS}$ ⟹ ΔABD ~ ΔPQS (SAS) ⟹ $\frac{AD}{PS}=\frac{AB}{PQ}$. Similarly ΔADC ~ ΔPSR (SAS), using $\angle C=\angle R$ and $\frac{DC}{SR}=\frac{AC}{PR}$.
(ii) From ΔABD ~ ΔPQS: $\frac{AD}{PS}=\frac{AB}{PQ}=\frac{BC}{QR}$. Hence $\frac{AD}{PS}=\frac{BC}{QR}$.
