English

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

Question

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

Solution

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
  Is there an error in this question or solution?
Chapter 13: Quadrilaterals - Exercise 13.4 [Page 63]

APPEARS IN

RD Sharma Mathematics [English] Class 9
Chapter 13 Quadrilaterals
Exercise 13.4 | Q 5 | Page 63

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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.


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)


ABCD is a quadrilateral in which AD = BC. E, F, G and H are the mid-points of AB, BD, CD and Ac respectively. Prove that EFGH is a rhombus.


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.


In the figure, give below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that:
(i) AQ // BS
(ii) DS = 3 Rs.


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.


If L and M are the mid-points of AB, and DC respectively of parallelogram ABCD. Prove that segment DL and BM trisect diagonal AC.


In a parallelogram ABCD, M is the mid-point AC. X and Y are the points on AB and DC respectively such that AX = CY. Prove that:
(i) Triangle AXM is congruent to triangle CYM, and

(ii) XMY is a straight line.


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: ∠EFG = 90°


In ∆ABC, AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are respectively the mid-points of AB and BC, determine the length of DE.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×