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
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.
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.
Fill in the blank to make the following statement correct
The triangle formed by joining the mid-points of the sides of an isosceles triangle is
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
In triangle ABC, AD is the median and DE, drawn parallel to side BA, meets AC at point E.
Show that BE is also a median.
In the given figure, AD and CE are medians and DF // CE.
Prove that: FB = `1/4` AB.
Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square is also a square.
In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: AB, if DC = 8 cm and PQ = 9.5 cm
E is the mid-point of a median AD of ∆ABC and BE is produced to meet AC at F. Show that AF = `1/3` AC.
