Advertisements
Advertisements
प्रश्न
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
Advertisements
उत्तर

Since ABCD is a parallelogram,
AB = CD and AD = BC
Now, E and F are the mid-points of AB and CD respectively,
⇒ AE = EB = DF = FC ....(i)
In ΔHEB and ΔHFC,
BE = FC ....[From (i)]
∠EHB = ∠FHC ....(vertically opposite angles)
∠HBE = ∠HFC ....(Alternate interior angles)
∴ ΔHEB ≅ ΔHFC.
APPEARS IN
संबंधित प्रश्न
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.

Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
In a ∆ABC, D, E and F are, respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of ∆DEF.
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.
In a triangle, P, Q and R are the mid-points of sides BC, CA and AB respectively. If AC =
21 cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ.
In below Fig, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = `1/4` AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.

In the given figure, M is mid-point of AB and DE, whereas N is mid-point of BC and DF.
Show that: EF = AC.
In trapezium ABCD, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC.
Prove that: AB + DC = 2EF.
In triangle ABC; M is mid-point of AB, N is mid-point of AC and D is any point in base BC. Use the intercept Theorem to show that MN bisects AD.
In the given figure, PS = 3RS. M is the midpoint of QR. If TR || MN || QP, then prove that:
RT = `(1)/(3)"PQ"`
