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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 5 · Set 3 · Q32
(a)D is the mid-point of side BC of $\displaystyle \triangle \mathrm{ABC}$. CE and BF intersect at O, a point on AD . AD is produced to G such that $\displaystyle \mathrm{OD}=\mathrm{DG}$. Prove that(i)OBGC is a parallelogram.(ii)$\displaystyle \mathrm{EF}|\mid \mathrm{BC}$(iii)$\displaystyle \triangle \mathrm{AEF} \sim \triangle \mathrm{ABC}$
(b)Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. Prove that(i)$\displaystyle \mathrm{AQ}=\mathrm{QR}$(ii)$\displaystyle \mathrm{AP}=2 \mathrm{PQ}$(iii)$\displaystyle \mathrm{PR}=2 \mathrm{AP}$
(a)
D is the mid-point of side BC of $\displaystyle \triangle \mathrm{ABC}$. CE and BF intersect at O, a point on AD . AD is produced to G such that $\displaystyle \mathrm{OD}=\mathrm{DG}$. Prove that
(i)
OBGC is a parallelogram.
(ii)
$\displaystyle \mathrm{EF}|\mid \mathrm{BC}$
(iii)
$\displaystyle \triangle \mathrm{AEF} \sim \triangle \mathrm{ABC}$
(b)
Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. Prove that
(i)
$\displaystyle \mathrm{AQ}=\mathrm{QR}$
(ii)
$\displaystyle \mathrm{AP}=2 \mathrm{PQ}$
(iii)
$\displaystyle \mathrm{PR}=2 \mathrm{AP}$
Marking-scheme solution
∵ Diagonals OG and BC of quadrilateral OBGC bisect each other.
∴ OBGC is a parallelogram
(ii) \(\displaystyle \mathrm{CO}\|\mathrm{GB} \Rightarrow \mathrm{CE}\| \mathrm{GB}\)
In \(\displaystyle \triangle \mathrm{AGB}, \mathrm{OE} \| \mathrm{GB} \Rightarrow \frac{\mathrm{AO}}{\mathrm{OG}}=\frac{\mathrm{AE}}{\mathrm{EB}}\)
Similarly in \(\displaystyle \triangle \mathrm{AGC}, \quad \frac{\mathrm{AO}}{\mathrm{OG}}=\frac{\mathrm{AF}}{\mathrm{FC}}\)
\(\displaystyle \Rightarrow \frac{\mathrm{AE}}{\mathrm{EB}}=\frac{\mathrm{AF}}{\mathrm{FC}} \Rightarrow \mathrm{EF} \| \mathrm{BC}\)
(iii) In \(\displaystyle \triangle \mathrm{AEF}\) and \(\displaystyle \triangle \mathrm{ABC}\)
\(\displaystyle \angle \mathrm{AEF}=\angle \mathrm{ABC}\) and \(\displaystyle \angle \mathrm{A}\) is common.
\(\displaystyle \therefore \triangle \mathrm{AEF} \sim \triangle \mathrm{ABC}\)
\(\displaystyle \mathrm{QC} \| \mathrm{AB} \therefore \triangle \mathrm{RQC} \sim \triangle \mathrm{RAB}\)
\[\begin{aligned}
& \Rightarrow \frac{Q R}{A R}=\frac{Q C}{A B}=\frac{1}{2} \\
& \Rightarrow 2 Q R=A R \Rightarrow Q \text { is the mid point of } A R
\end{aligned}
\]
(ii) \(\displaystyle \triangle \mathrm{PQD} \sim \triangle \mathrm{PAB}\)
\(\displaystyle \therefore \frac{\mathrm{QP}}{\mathrm{AP}}=\frac{\mathrm{DQ}}{\mathrm{BA}}=\frac{1}{2}\)
\(\displaystyle \Rightarrow \mathrm{AP}=2 \mathrm{PQ}\)
(iii) Since \(\displaystyle \mathrm{AQ}=\mathrm{QR}\)
\(\displaystyle \Rightarrow \mathrm{AP}+\mathrm{PQ}=\mathrm{PR}-\mathrm{PQ}\)
\(\displaystyle \Rightarrow \mathrm{AP}+\frac{1}{2} \mathrm{AP}=\mathrm{PR}-\frac{1}{2} \mathrm{AP}\)
\(\displaystyle \Rightarrow \mathrm{PR}=2 \mathrm{AP}\)
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CBSE Class 10 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.