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 OD = DG. 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) AQ = QR
(ii) AP = 2PQ
(iii) PR = 2AP

Approach

In $△ A BC$ , D is the midpoint of BC. CE and BF intersect at O on AD, and AD is produced to G such that OD = DG. (i) Show OBGC is a parallelogram: Since diagonals OG and BC bisect each other at D, OBGC is a parallelogram. (ii) Show EF || BC: In $△ A GB$ , since OE || GB (from OBGC being a parallelogram, CO || GB implies CE || GB), by BPT, $\frac{A O}{OG} = \frac{A E}{EB}$ . Similarly in $△ A GC$ , $\frac{A O}{OG} = \frac{A F}{FC}$ . Hence, $\frac{A E}{EB} = \frac{A F}{FC}$ , so EF || BC. (iii) Show $△ A EF \sim △ A BC$ : Since EF || BC, $∠ A EF = ∠ A BC$ (corresponding angles). $∠ A$ is common. By AA similarity, $△ A EF \sim △ A BC$ .

Step-by-step working

1
(i) Diagonals OG and BC of quadrilateral OBGC bisect each other at D (given OD = DG and BD = DC). Hence, OBGC is a parallelogram.
1 mark
2
(ii) Since OBGC is a parallelogram, CO || GB. In $△ A GB$ , OE || GB, so by BPT, $\frac{A O}{OG} = \frac{A E}{EB}$ .
0.5 marks
3
Similarly, in $△ A GC$ , OF || GC (since CO || GB implies OF || GC), so $\frac{A O}{OG} = \frac{A F}{FC}$ .
1 mark
4
Hence, $\frac{A E}{EB} = \frac{A F}{FC}$ . By the converse of BPT, EF || BC.
0.5 marks
5
(iii) In $△ A EF$ and $△ A BC$ , $∠ A EF = ∠ A BC$ (corresponding angles since EF || BC) and $∠ A$ is common.
1 mark
6
By AA similarity, $△ A EF \sim △ A BC$ .
1 mark

Concepts used

trianglesparallelogrammidpoint theoremsimilar trianglesprove

Approach

Step-by-step working

Concepts used

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