Question

(a) D is the mid-point of side BC of $△ A BC$ . CE and BF intersect at O, a point on AD. AD is produced to G such that $O D = D G$ . Prove that
(i) OBGC is a parallelogram.
(ii) $EF ∥ BC$
(iii) $△ A EF \sim △ A BC$

(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) $A Q = QR$
(ii) $A P = 2 PQ$
(iii) $PR = 2 A P$

Approach

In triangle ABC, D is midpoint of BC. CE and BF intersect at O on AD, with AD produced to G such that OD = DG. Prove: (i) OBGC is a parallelogram (diagonals bisect each other). (ii) EF is parallel to BC (by midpoint theorem applied twice). (iii) Triangle AEF is similar to triangle ABC (AA similarity).

Step-by-step working

1
(i) Diagonals OG and BC of quadrilateral OBGC bisect each other at D, hence OBGC is a parallelogram.
1 mark
2
(ii) In parallelogram OBGC, $CO ∥ GB$ , thus $CE ∥ GB$ .
0.5 marks
3
In $△ A GB$ , since OE is parallel to GB by midpoint theorem, $\frac{A O}{OG} = \frac{A E}{EB}$ .
1 mark
4
Similarly in $△ A GC$ , $\frac{A O}{OG} = \frac{A F}{FC}$ . Thus $\frac{A E}{EB} = \frac{A F}{FC}$ , so $EF ∥ BC$ by converse of BPT.
0.5 marks
5
(iii) In $△ A EF$ and $△ A BC$ , $∠ A EF = ∠ A BC$ (corresponding angles, $EF ∥ BC$ ) and $∠ A$ is common.
1 mark
6
Hence $△ A EF \sim △ A BC$ by AA similarity.
1 mark

Concepts used

parallelogrammid-point theoremsimilar trianglesprove

Approach

Step-by-step working

Concepts used

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