3/6/2024 0 Comments In parallelogram abcd![]() Show that:(i) ABCD is a square(ii) diagonal BD bisects ∠B as well as ∠D. Given : Diagonals AC and BD of a quadrilateral ABCD intersect. The next step is to visualize the sides of the parallelogram so you can determine. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. ABCD intersect at O in such a way that ar(AOD) ar(BOC). of parallelogram ABCD on the coordinate grid provided with the question.Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D. ![]() Show that i) it bisects ∠C also, ii) ABCD is a rhombus. Diagonal AC of a parallelogram ABCD bisects ∠A (see the given figure).Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.If in parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ, then ΔAPD ≅ ΔCQB by SAS congruence, AP = CQ, ΔAQB ≅ ΔCPD by SAS congruence, AQ = CP, and APCQ is a parallelogram. In a parallelogram ABCD, it is being given that AB 10 cm and the altitudes corresponding to the sides AB and AD are DL 6 cm and BM 8 cm, respectively. NCERT Maths Solutions Class 9 Chapter 8 Exercise 8.1 Question 9 Prove that : i AXYC ii AX is parallel to YC iii AXCY is a parallelogram. Show that: (i) ΔAPD ≅ ΔCQB (ii) AP = CQ (iii) ΔAQB ≅ ΔCPD (iv) AQ = CP (v) APCQ is a parallelogram In parallelogram ABCD, X and Y are mid points of opposite sides AB and DC respectively. Video Solution: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. Let D be the midpoint of O Adot using vector methods prove that B Da asked in Algebra by JohnAgrawal ( 91. Let G be the intersection of the diagonal DB and the line segment EF. Let O A C B be a parallelogram with O at the origin andO C a diagonal. ☛ Check: NCERT Solutions Class 9 Maths Chapter 8 Question 1136360: In parallelogram ABCD, E is the midpoint of AB and F is the midpoint of DC. Since opposite sides in quadrilateral APCQ are equal to each other, thus APCQ is a parallelogram. (v) From the result obtained in (ii) and (iv), AQ = CP and AP = CQ ∴ ΔAQB ≅ ΔCPD (Using SAS congruence rule) ∠ABQ = ∠CDP ( Alternate interior angles for AB || CD) ∴ ΔAPD ≅ ΔCQB (Using SAS congruence rule)ĪB = CD (Opposite sides of parallelogram ABCD) In triangle ADE, The height of the parallelogram is 8.632. It means the base of parallelogram is 11.14 in. According to the angle sum property, the sum of interior angles of a triangle is 180º. ∠ADP = ∠CBQ (Alternate interior angles for BC || AD)ĪD = CB (Opposite sides of parallelogram ABCD) Opposite angles of a parallelogram are congruent. Given: ABCD is a parallelogram and DP = BQ In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig.
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