हिंदी

In a δAbc, E and F Are the Mid-points of Ac and Ab Respectively. the Altitude Ap to Bc Intersects Fe at Q. Prove that Aq = Qp. - Mathematics

Advertisements
Advertisements

प्रश्न

In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC
intersects FE at Q. Prove that AQ = QP.

Advertisements

उत्तर

In ΔABC

E and F are midpoints of AB and  AC

∴ EF || FE, `1/2` BC =FE                [  ∴ By mid-point theorem]

In ΔABP

F is the midpoint of AB and   FQ || BP         [ ∵ EF || BC ]

∴ Q is the midpoint of AP                     [By converse of midpoint theorem]

Hence, AQ = QP

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 13: Quadrilaterals - Exercise 13.4 [पृष्ठ ६३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 9
अध्याय 13 Quadrilaterals
Exercise 13.4 | Q 5 | पृष्ठ ६३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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.


ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the
mid-points of the sides, in order, is a rectangle.


The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:

PR = `[1]/[2]` ( AB + CD)


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 parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.


In the given figure, AD and CE are medians and DF // CE.
Prove that: FB = `1/4` AB.


If the quadrilateral formed by joining the mid-points of the adjacent sides of quadrilateral ABCD is a rectangle,
show that the diagonals AC and BD intersect at the right angle.


In ΔABC, BE and CF are medians. P is a point on BE produced such that BE = EP and Q is a point on CF produced such that CF = FQ. Prove that: A is the mid-point of PQ.


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: AP = 2AR


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×