1If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Answer & Reference:
In a parallelogram, diagonals bisect each other (Theorem 8.6). If they are also equal, then the triangles formed are congruent by SSS, making all angles 90°. Hence, it is a rectangle. (Implied from Example 1 and Theorem 8.1)
2Show that the diagonals of a square are equal and bisect each other at right angles.
Answer & Reference:
A square is both a rectangle (diagonals equal) and a rhombus (diagonals perpendicular). From Theorem 8.6, diagonals bisect each other. From Example 2, diagonals of a rhombus are perpendicular. Hence, diagonals of a square are equal, bisect each other, and intersect at 90°. (Summary point 5)
3Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.11). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.
Answer & Reference:
(i) Since ABCD is a parallelogram, ∠DAC = ∠BCA and ∠BAC = ∠DCA (alternate angles). If AC bisects ∠A, then ∠DAC = ∠BAC ⇒ ∠BCA = ∠DCA, so AC bisects ∠C.
(ii) Then △ABC ≅ △ADC (ASA), so AB = BC. Hence all sides equal → rhombus. (Based on Theorem 8.1 and properties of parallelogram)
Fig. 8.11: Parallelogram ABCD with diagonal AC
4ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that: (i) ABCD is a square, (ii) diagonal BD bisects ∠B as well as ∠D.
Answer & Reference:
(i) In rectangle ABCD, if AC bisects ∠A (90°), then ∠BAC = 45°. In △ABC, ∠ABC = 90°, so ∠BCA = 45° ⇒ AB = BC. Adjacent sides equal → square.
(ii) In a square, diagonals bisect angles (property of rhombus + rectangle). So BD bisects ∠B and ∠D. (From Example 2 and rectangle properties)
5In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.12). Show that: (i) △APD ≅ △CQB, (ii) AP = CQ, (iii) △AQB ≅ △CPD, (iv) AQ = CP, (v) APCQ is a parallelogram.
Answer & Reference:
Using alternate angles (∠ADP = ∠CBQ) and AD = BC (Theorem 8.2), and DP = BQ (given), △APD ≅ △CQB (SAS). So AP = CQ (CPCT). Similarly, △AQB ≅ △CPD ⇒ AQ = CP. Since both pairs of opposite sides equal, APCQ is a parallelogram. (Based on Theorem 8.2 and congruence)
Fig. 8.12: Points P and Q on diagonal BD
6ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.13). Show that (i) △APB ≅ △CQD, (ii) AP = CQ.
Answer & Reference:
∠ABP = ∠CDQ (alternate angles), ∠APB = ∠CQD = 90°, and AB = CD (Theorem 8.2). So △APB ≅ △CQD (AAS). Hence AP = CQ (CPCT). (Based on Theorem 8.2 and angle properties)
Fig. 8.13: Perpendiculars AP and CQ on BD
7ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.14). Show that (i) ∠A = ∠B, (ii) ∠C = ∠D, (iii) △ABC ≅ △BAD, (iv) diagonal AC = diagonal BD.
Answer & Reference:
Extend AB and draw CE || DA. Then AECD is a parallelogram ⇒ AD = CE. But AD = BC ⇒ BC = CE ⇒ ∠CEB = ∠CBE. Using linear pair and parallel lines, ∠A = ∠B. Similarly, ∠C = ∠D. Then △ABC ≅ △BAD (SAS), so AC = BD. (As per given Hint and Theorem 8.2)
Fig. 8.14: Isosceles trapezium ABCD
8ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.20). AC is a diagonal. Show that: (i) SR || AC and SR = ½ AC, (ii) PQ = SR, (iii) PQRS is a parallelogram.
Answer & Reference:
By Mid-point Theorem (Theorem 8.8), in △ADC, S and R mid-points ⇒ SR || AC and SR = ½ AC. Similarly, in △ABC, PQ || AC and PQ = ½ AC. So PQ = SR and PQ || SR ⇒ PQRS is a parallelogram.
Fig. 8.20: Mid-points P, Q, R, S
9ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Answer & Reference:
Join diagonals. In rhombus, diagonals are perpendicular. Using Mid-point Theorem, PQRS has opposite sides parallel and equal. Also, one angle = 90° (since diagonals ⊥), so PQRS is a rectangle. (Similar to Example 5 logic)
10ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Answer & Reference:
In rectangle, diagonals are equal. By Mid-point Theorem, all sides of PQRS = ½ diagonal ⇒ all sides equal. Also, opposite sides parallel ⇒ rhombus. (Based on Theorem 8.8 and rectangle properties)
11ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.21). Show that F is the mid-point of BC.
Answer & Reference:
In △ABD, E is mid-point of AD and EF || AB ⇒ by converse of Mid-point Theorem (Theorem 8.9), the line through E || AB meets BD at its mid-point. Then in △BDC, same logic ⇒ F is mid-point of BC.
Fig. 8.21: E mid-point, EF || AB
12In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.22). Show that the line segments AF and EC trisect the diagonal BD.
Answer & Reference:
Since AB || CD and AB = CD, AE = FC and AE || FC ⇒ AECF is a parallelogram. So AF || EC. In △DQC and △APB, using Theorem 8.9, the intersections divide BD into three equal parts. Hence, AF and EC trisect BD.
Fig. 8.22: Mid-points E and F
13ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC, (ii) MD ⊥ AC, (iii) CM = MA = ½ AB.
Answer & Reference:
(i) By Theorem 8.9 (converse of mid-point theorem), since M is mid-point of AB and MD || BC, D is mid-point of AC.
(ii) Since BC ⊥ AC and MD || BC ⇒ MD ⊥ AC.
(iii) In △AMD and △CMD, AD = CD, MD common, ∠MDA = ∠MDC = 90° ⇒ △AMD ≅ △CMD ⇒ MA = CM. But M is mid-point of AB ⇒ MA = ½ AB. So CM = MA = ½ AB.