NCERT Solutions

State whether the following quadrilaterals are similar or not: (The first quadrilateral is a rhombus PQRS with side length 1.5 cm and corner angles that are not 90 degrees. The second quadrilateral is a square ABCD with side length 3 cm and corner angles of 90 degrees.)

In a figure, if $LM \| CB$ and $LN \| CD$, prove that $\frac{AM}{AB} = \frac{AN}{AD}$. The figure shows a triangle ABC and a triangle ADC, sharing the side AC.

In a figure, $DE \| AC$ and $DF \| AE$. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$. The figure shows a triangle ABC, with point D on AB, E on BC, and F on BE.

In a figure, $DE \| OQ$ and $DF \| OR$. Show that $EF \| QR$. The figure shows a triangle PQR with a point O inside it. Lines OP, OQ, OR are drawn. D is on PQ, E is on PR, F is on QR. No, D is on OP, E is on OQ, F is on OR. Let's re-read the text. The figure shows a triangle PQR with point O inside. D is a point on PQ, E is a point on PR. The text seems to contradict the figure description. Let me follow the text: In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR. Figure 6.20 shows a triangle PQR, a point O. Lines are drawn from O to P, Q, R. D is on PQ, E is on PR... wait, the figure from the source shows D on PQ, E is on PR... No, the figure shows D is on PO, E is on PQ and F is on PR. Let me re-read the figure description from the text again. 'In Fig. 6.20, DE || OQ and DF || OR'. The figure shows point O, and vertices P, Q, R. Lines OP, OQ, OR are drawn. A triangle DEF is formed with D on PQ, E is on PR... this is getting more confusing. Let's assume a standard configuration for this problem type. P is a vertex, PO, PQ, PR are lines from P. O is some other point. Let's assume the question meant a configuration like in Q6. Let's re-read the source for Q5. In Fig. 6.20, DE || OQ and DF || OR. Fig 6.20 shows a triangle PQR, and a point O inside. Lines PO, QO, RO are not drawn. Instead, lines OP, OQ, OR are drawn from vertex O to P, Q, R. A triangle DEF is drawn inside such that D is on PQ, E is on PR... this is getting more confusing. Let's assume the standard problem setup: In $\triangle PQR$, O is a point in its interior. D is on PQ, E is on QR, F is on PR. Then the question would be different. Let's assume the setup of Fig 6.21 is intended for Q5 and Q6. A point O outside the triangle, with P,Q,R collinear with points A,B,C. No, that's for Q6. Let's re-examine Fig 6.20. It shows a point P, and rays PQ and PR forming an angle. O is a point. Lines OQ and OR are drawn. E is on PQ, F is on PR. DE is parallel to OQ, DF is parallel to OR. D must be a point on OP. Yes, this makes sense. D is on OP, E is on PQ, F is on PR.

In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that $AB \| PQ$ and $AC \| PR$. Show that $BC \| QR$.