हिंदी

Bm and Cn Are Perpendiculars to a Line Passing Through the Vertex a of a Triangle Abc. If L is the Mid-point of Bc, Prove that Lm = Ln. - Mathematics

Advertisements
Advertisements

प्रश्न

BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If
L is the mid-point of BC, prove that LM = LN.

Advertisements

उत्तर

To prove LM = LN
Draw LS perpendicular to line MN

∴ The lines BM, LS and CN being the same perpendiculars, on line MN are parallel to each
other.
According to intercept theorem,
If there are three or more parallel lines and the intercepts made by them on a transversal or
equal. Then the corresponding intercepts on any other transversal are also equal.
In the drawn figure, MB and LS and NC are three parallel lines and the two transversal line
are MN and BC
We have, BL= LC  (As L is the given midpoint of BC)
∴ using intercept theorem, we get

MS = SN               ....(i )

Now in Δ MLS and LSN

MS = SN using                ….(i)

`∠`LSM = `∠`LSN = 90°LS ^ MN and SL = LS common

∴ Δ DMLS ≅  Δ LSN       (SAS congruency theorem)

∴ LM = LN   (CPCT )

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 13: Quadrilaterals - Exercise 13.4 [पृष्ठ ६५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 9
अध्याय 13 Quadrilaterals
Exercise 13.4 | Q 20 | पृष्ठ ६५

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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 a triangle ABC, AD is a median and E is mid-point of median AD. A line through B and E meets AC at point F.

Prove that: AC = 3AF.


In triangle ABC, P is the mid-point of side 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 : (i) AP = 2AR
                   (ii) BC = 4QR


Use the following figure to find:
(i) BC, if AB = 7.2 cm.
(ii) GE, if FE = 4 cm.
(iii) AE, if BD = 4.1 cm
(iv) DF, if CG = 11 cm.


Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square is also a square.


Side AC of a ABC is produced to point E so that CE = `(1)/(2)"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 meets AC at point P and EF at point R respectively. Prove that: 3DF = EF


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.


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: ΔHEB ≅ ΔHFC


In ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: Q A and P are collinear.


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]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×