Advertisements
Advertisements
Question
ABC is a triang D is a point on AB such that AD = `1/4` AB and E is a point on AC such that AE = `1/4` AC. Prove that DE = `1/4` BC.
Advertisements
Solution
Let P and Q be the midpoints of AB and AC respectively.
Then PQ || BC such that
PQ = `1/2` BC ......(i)
In ΔAPQ, D and E are the midpoint of AP and AQ are respectively
∴ DE || PQ and DE = `1/2` PQ ....(ii)
From (1) and (2) DE = `1/2 PQ = 1/2 PQ = 1/2 (1/2 BC) `
DE = `1 /4`BC
Hence, proved.
APPEARS IN
RELATED QUESTIONS
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`
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 Fig. below, M, N and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC.
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
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.
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
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 ΔABC, D, E and F are the midpoints of AB, BC and AC.
Show that AE and DF bisect each other.
P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.
Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.
