Advertisements
Advertisements
Question
D, E and F are the mid-points of the sides AB, BC and CA of an isosceles ΔABC in which AB = BC. Prove that ΔDEF is also isosceles.
Advertisements
Solution
E and F are mid-points of BC and AC
Therefore, EF = `(1)/(2)"AB"`......(i)
D and F are mid-points of AB and AC
Therefore, DF = `(1)/(2)"BC"`......(ii)
But AB = BC
From (i) and (ii)
EF = DF
Therefore, ΔDEF is an isosceles triangle.
APPEARS IN
RELATED QUESTIONS
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 below Fig, ABCD and PQRC are rectangles and Q is the mid-point of Prove thaT
i) DP = PC (ii) PR = `1/2` AC
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral
bisect each other.
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 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"`.
ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: ∠EFG = 90°
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: ΔGEA ≅ ΔGFD
E is the mid-point of the side AD of the trapezium ABCD with AB || DC. A line through E drawn parallel to AB intersect BC at F. Show that F is the mid-point of BC. [Hint: Join AC]
D, E and F are the mid-points of the sides BC, CA and AB, respectively of an equilateral triangle ABC. Show that ∆DEF is also an equilateral triangle.
P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square.
