Advertisements
Advertisements
प्रश्न
ΔABC is an isosceles triangle with AB = AC. D, E and F are the mid-points of BC, AB and AC respectively. Prove that the line segment AD is perpendicular to EF and is bisected by it.
Advertisements
उत्तर
Since the segment joining the mid-points of two sides of a triangle is parallel to third side and is half of it,
Therefore,
DE || AB, DE = `(1)/(2)"AB"`
Also,
DF || AC, DF = `(1)/(2)"AC"`
But AB = AC
⇒ `(1)/(2)"AB" = (1)/(2)"AC"`
⇒ DF = DE ........(i)
DE = `(1)/(2)"AB"`
⇒ DE = AF ........(ii)
And DF = `(1)/(2)"AC"`
⇒ DF = AE ........(iii)
From (i), (ii) and (iii)
DE = AE = EF = DF
⇒ DEAF is a rhombus.
⇒ Diagonals AD and EF bisect each other at right angles.
⇒ AD perpendicular to EF and AD is bisected by EF.
APPEARS IN
संबंधित 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.
In the adjacent figure, `square`ABCD is a trapezium AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.
The side AC of a triangle ABC is produced to point E so that CE = AC. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively.
Prove that:
- 3DF = EF
- 4CR = AB
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 triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a parallelogram.
In a right-angled triangle ABC. ∠ABC = 90° and D is the midpoint of AC. Prove that BD = `(1)/(2)"AC"`.
In ΔABC, P is the mid-point of BC. A line through P and parallel to CA meets AB at point Q, and a line through Q and parallel to BC meets median AP at point R. Prove that: BC = 4QR
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: F is the mid-point of BC.
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.
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.
