Advertisements
Advertisements
प्रश्न
In the given figure, ΔABC is an equilateral traingle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ΔFED is an equilateral traingle.

Advertisements
उत्तर
Given: ∆ABC is an equilateral triangle and D, E and F are mid-points of BC, AC and AB respectively.
To prove: ∆FED is an equilateral triangle.
Proof:
In ΔABC,
Points F and E are the midpoints of sides AB and AC respectively. ...(Given)
∴ FE = `1/2` BC ...(From midpoint theorem) ...(i)
In ΔABC,
Points D and E are the midpoints of sides BC and AC respectively. ...(Given)
∴ DE = `1/2` AB ...(From midpoint theorem) ...(ii)
In ΔABC,
Points D and F are the midpoints of sides BC and AB respectively. ...(Given)
∴ DF = `1/2` AC ...(From midpoint theorem) ...(iii)
Now, ΔABC is an equilateral triangle.
∴ BC = AB = AC ...(Sides of equilateral triangle)
∴ `1/2` BC = `1/2` AB = `1/2` AC ...(Multiplying both sides by `1 /2`)
∴ FE = DE = DF ...[From (i), (ii) and (iii)]
∴ ΔFED is an equilateral triangle.
APPEARS IN
संबंधित प्रश्न
ABC 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
- D is the mid-point of AC
- MD ⊥ AC
- CM = MA = `1/2AB`
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 Fig. below, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are
respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.

ABCD is a parallelogram, E and F are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GP = PH.
Fill in the blank to make the following statement correct:
The triangle formed by joining the mid-points of the sides of a right triangle is
In triangle ABC, M is mid-point of AB and a straight line through M and parallel to BC cuts AC in N. Find the lengths of AN and MN if Bc = 7 cm and Ac = 5 cm.
In a triangle ABC, AD is a median and E is mid-point of median AD. A line through B and E meets AC at point F.
Prove that: AC = 3AF.
A parallelogram ABCD has P the mid-point of Dc and Q a point of Ac such that
CQ = `[1]/[4]`AC. PQ produced meets BC at R.

Prove that
(i)R is the midpoint of BC
(ii) PR = `[1]/[2]` DB
D, E, and F are the mid-points of the sides AB, BC, and CA respectively of ΔABC. AE meets DF at O. P and Q are the mid-points of OB and OC respectively. Prove that DPQF is a parallelogram.
In triangle ABC, the medians BP and CQ are produced up to points M and N respectively such that BP = PM and CQ = QN. Prove that:
- M, A, and N are collinear.
- A is the mid-point of MN.
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 parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.
In ΔABC, BE and CF are medians. P is a point on BE produced such that BE = EP and Q is a point on CF produced such that CF = FQ. Prove that: A is the mid-point of PQ.
In a right-angled triangle ABC. ∠ABC = 90° and D is the midpoint of AC. Prove that BD = `(1)/(2)"AC"`.
In parallelogram ABCD, P is the mid-point of DC. Q is a point on AC such that CQ = `(1)/(4)"AC"`. PQ produced meets BC at R. Prove that
(i) R is the mid-point of BC, and
(ii) PR = `(1)/(2)"DB"`.
Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square is also a square.
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.
In the given figure, PS = 3RS. M is the midpoint of QR. If TR || MN || QP, then prove that:
ST = `(1)/(3)"LS"`
D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. Prove that by joining these mid-points D, E and F, the triangles ABC is divided into four congruent triangles.
