Advertisements
Advertisements
प्रश्न
The diagonals of a quadrilateral intersect each other at right angle. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is a rectangle.
Advertisements
उत्तर
P and Q are mid-points of AB and BC.
∴ PQ || AC and PQ = `(1)/(2)"AC"`.......(i)
S and R are mid-points of AD and DC.
∴ SR || AC and SR = `(1)/(2)"AC"`.......(ii)
From (i) and (ii)
PQ || SR and PQ = SR
Therefore, PQRS is a parallelogram.
Further AC and BD intersect at right angles
∴ SP || BD and BD ⊥ AC
∴ SP ⊥ AC
⇒ SP ⊥ SR
⇒ ∠RSP = 90°
∴ ∠ RSP = ∠SRQ = ∠RQS = ∠SPQ = 90°
Therefore, PQRS is a rectangle.
APPEARS IN
संबंधित प्रश्न
ABCD 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.
ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.
In the figure, give below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that:
(i) AQ // BS
(ii) DS = 3 Rs.
In parallelogram ABCD, E is the mid-point of AB and AP is parallel to EC which meets DC at point O and BC produced at P.
Prove that:
(i) BP = 2AD
(ii) O is the mid-point of AP.
In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meets side BC at points M and N respectively. Prove that: BM = MN = NC.
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: DC, if AB = 20 cm and PQ = 14 cm
In AABC, D and E are two points on the side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet the side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meet the side BC at points M and N respectively. Prove that BM = MN = NC.
In the given figure, PS = 3RS. M is the midpoint of QR. If TR || MN || QP, then prove that:
ST = `(1)/(3)"LS"`
E is the mid-point of a median AD of ∆ABC and BE is produced to meet AC at F. Show that AF = `1/3` AC.
