Advertisements
Advertisements
प्रश्न
The diagonals AC and BD of a quadrilateral ABCD intersect at right angles. Prove that the quadrilateral formed by joining the midpoints of quadrilateral ABCD is a rectangle.
Advertisements
उत्तर
The figure is as below:
Let ABCD be a quadrilateral where p, Q, R, S are the midpoint of sides AB, BC, CD, DA respectively. Diagonals AC and BD intersect at right angles at point O. We need to show that PQRS is a rectangle
Proof:
In ΔABC and ΔADC,
2PQ = AC and PQ || AC ....(1)
2RS = AC and RS || AC ....(2)
From (1) and (2), we get
PQ = RS and PQ || RS
Similarly, we can show that
PS = RQ and PS || RQ
Therefore, PQRS is a parallelogram.
Now, PQ || AC,
∴ ∠AOD = ∠PXO = 90° ....(Corresponding angles)
Again, BD || RQ,
∴ ∠PXO = ∠RQX = 90° ....(Corresponding angles)
Similarly, ∠QRS = ∠RSP = SPQ = 90°
Therefore, PQRD is a rectangle.
APPEARS IN
संबंधित प्रश्न
In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of
the triangle formed by joining the mid-points of the sides of this triangle.
ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the
mid-points of the sides, in order, is a rectangle.
Use the following figure to find:
(i) BC, if AB = 7.2 cm.
(ii) GE, if FE = 4 cm.
(iii) AE, if BD = 4.1 cm
(iv) DF, if CG = 11 cm.

In triangle ABC ; D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F.
Prove that BDEF is a parallelogram. If AB = 16 cm, AC = 12 cm and BC = 18 cm,
find the perimeter of the parallelogram BDEF.
In ΔABC, D is the mid-point of AB and E is the mid-point of BC.
Calculate:
(i) DE, if AC = 8.6 cm
(ii) ∠DEB, if ∠ACB = 72°
In a right-angled triangle ABC. ∠ABC = 90° and D is the midpoint of AC. Prove that BD = `(1)/(2)"AC"`.
ABCD is a parallelogram.E is the mid-point of CD and P is a point on AC such that PC = `(1)/(4)"AC"`. EP produced meets BC at F. Prove that: 2EF = BD.
In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: ΔHEB ≅ ΔHFC
In the given figure, PS = 3RS. M is the midpoint of QR. If TR || MN || QP, then prove that:
RT = `(1)/(3)"PQ"`
The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is ______.
