Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
The figure is shown below

(i) In ΔAQN & ΔBQC
AQ = QB (Given)
∠AQN = ∠BQC
QN = QC
∴ ΔAQN ≅ ΔBQC ...[ by SAS ]
∴ ∠QAN = ∠QBC ...(1)
And BC = AN ……(2)
Similarly, ΔAPM ≅ ΔCPB .....[by SAS]
∠PAM = ∠PCB ...(3) [by CPTC]
And BC = AM ….( 4 )
Now In ΔABC,
∠ABC + ∠ACB + ∠BAC = 180°
∠QAN + ∠PAM + ∠BAC = 180° ...[ (1), (2) we get ]
Therefore M, A, N are collinear.
(ii) From (3) and (4) MA = NA
Hence A is the midpoint of MN.
APPEARS IN
संबंधित प्रश्न
Fill in the blank to make the following statement correct:
The figure formed by joining the mid-points of consecutive sides of a quadrilateral is
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.
In Δ ABC, AD is the median and DE is parallel to BA, where E is a point in AC. Prove that BE is also a median.
In ΔABC, D, E, F are the midpoints of BC, CA and AB respectively. Find ∠FDB if ∠ACB = 115°.
In a right-angled triangle ABC. ∠ABC = 90° and D is the midpoint of AC. Prove that BD = `(1)/(2)"AC"`.
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.
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: The line drawn through G and parallel to FE and bisects DA.
In ΔABC, D, E and F are the midpoints of AB, BC and AC.
Show that AE and DF bisect each other.
In the given figure, T is the midpoint of QR. Side PR of ΔPQR is extended to S such that R divides PS in the ratio 2:1. TV and WR are drawn parallel to PQ. Prove that T divides SU in the ratio 2:1 and WR = `(1)/(4)"PQ"`.
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]
