Premium Online Home Tutors

3 Tutor System

Starting just at 265/hour

Show that

(i) \(\triangle{AMC}\) \(\displaystyle \cong \) \(\triangle{BMD}\)

(ii) \(\angle{DBC}\) is a right angle

(iii) \(\triangle{DBC}\) \(\displaystyle \cong \) \(\triangle{ACB}\)

(iv)CM = (1/2) AB

Answer :

Given :

\(\angle{ACB}\) in which \(\angle{C}\) = \(90^\circ\) and M is the mid-point of AB.

To prove :

(i) \(\triangle{AMC}\) \(\displaystyle \cong \) \(\triangle{BMD}\)

(ii) \(\angle{DBC}\) is a right angle

(iii) \(\triangle{DBC}\) \(\displaystyle \cong \) \(\triangle{ACB}\)

(iv)CM = (\(\frac{1}{2} \) ) AB

Construction : Produce CM to D, such that CM = MD. Join DB.

Proof : In \(\triangle{AMC}\) and \(\triangle{BMD}\), we have

AM = BM ...(M is the mid-point of AB)

CM = DM ...(Given)

and \(\angle{AMC}\) = \(\angle{BMD}\) ...(Vertically opposite angles)

\(\therefore \) \(\triangle{AMC}\) \(\displaystyle \cong \) \(\triangle{BMD}\)(By SAS congruency test)

Hence, part(i) is proved.

Also, AC = DB ...(By CPCT)

\(\angle{1}\) = \(\angle{2}\) ...(Alternate angles) and (by CPCT)

\(\therefore \) BD || CA and BC is transversal.

\(\Rightarrow \) \(\angle{ACB}\) + \(\angle{DBC}\) = \(180^\circ\)

But, \(\angle{ACB}\) = \(90^\circ\) ...(given)

\(\Rightarrow \) \(90^\circ\) + \(\angle{DBC}\) = \(180^\circ\)

\(\Rightarrow \) \(\angle{DBC}\) = \(180^\circ\) - \(90^\circ\)

\(\Rightarrow \) \(\angle{DBC}\) = \(90^\circ\)

Hence, part(ii) is proved, too.

Now, considering \(\triangle{DBC}\) and \(\triangle{ACB}\), we have,

AC = DB ...(from part(i))

Side BC is common.

and \(\angle{DBC}\) = \(\angle{ACB}\) = \(90^\circ\)

Therefore, \(\triangle{DBC}\) \(\displaystyle \cong \) \(\triangle{ACB}\) ...(SSA Congruency theorem)

Hence, now, part(iii) is proved, too.

Now, DC = AB ...(By CPCT)

Multipling both sides by \(\frac{1}{2} \) , we get,

(\(\frac{1}{2} \) ) DC = \(\frac{1}{2} \) ) AB

Now, as we know, CM = (\(\frac{1}{2} \) ) DC

Therefore, CM = (\(\frac{1}{2} \) ) AB

Hence, part (iv) is proved.

- In quadrilateral ACBD, AC = AD and AB bisects \(\angle{A}\) (see figure). Show that \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{ABD}\). What can you say about BC and BD ?
- ABCD is a quadrilateral in which AD = BC and \(\angle{DAB}\) = \(\angle{CBA}\) (see figure). Prove that (i) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{BAC}\) (ii) BD = AC (iii) \(\angle{ABD}\) = \(\angle{BAC}\)
- AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.
- l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{CDA}\).
- Line l is the bisector of a \(\angle{A}\) and \(\angle{B}\) is any point on l. BP and BQ are perpendiculars from B to the arms of \(\angle{A}\) (see figure). show that(i) \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{AQB}\)(ii) BP = BQ or B is equidistant from the arms of \(\angle{A}\).
- In figure, AC = AE, AB = AD and \(\angle{BAD}\) = \(\angle{EAC}\). Show that BC = DE.
- AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \(\angle{BAD}\) = \(\angle{ABE}\) and \(\angle{EPA}\) = \(\angle{DPB}\)(see figure). Show that (i) \(\triangle{DAP}\) \(\displaystyle \cong \) \(\triangle{EBP}\) (ii) AD = BE

- NCERT solutions for class 9 maths chapter 1 Number Systems
- NCERT solutions for class 9 maths chapter 2 Polynomials
- NCERT solutions for class 9 maths chapter 3 Coordinate geometry
- NCERT solutions for class 9 maths chapter 4 Linear equations in two variables
- NCERT solutions for class 9 maths chapter 5 Introduction to Euclidean Geometry
- NCERT solutions for class 9 maths chapter 6 Lines and Angles
- NCERT solutions for class 9 maths chapter 7 Triangles
- NCERT solutions for class 9 maths chapter 8 Quadrilaterals
- NCERT solutions for class 9 maths chapter 9 Areas of parallelograms and triangles
- NCERT solutions for class 9 maths chapter 10 Circles
- NCERT solutions for class 9 maths chapter 11 Constructions
- NCERT solutions for class 9 maths chapter 12 Heron's Formula
- NCERT solutions for class 9 maths chapter 13 Surface areas and volumes
- NCERT solutions for class 9 maths chapter 14 Statistics
- NCERT solutions for class 9 maths chapter 15 Probability

- NCERT solutions for class 9 science chapter 1 Matter in our Surroundings
- NCERT solutions for class 9 science chapter 2 Is Matter Around Us Pure
- NCERT solutions for class 9 science chapter 3 Atoms and Molecules
- NCERT solutions for class 9 science chapter 4 Structure of the Atom
- NCERT solutions for class 9 science chapter 5 The Fundamental Unit of Life
- NCERT solutions for class 9 science chapter 6 Tissues and Fundamental unit of life
- NCERT solutions for class 9 science chapter 7 Diversity in Living Organisms
- NCERT solutions for class 9 science chapter 8 Motion
- NCERT solutions for class 9 science chapter 9 Force and Laws of Motion
- NCERT solutions for class 9 science chapter 10 Gravitation
- NCERT solutions for class 9 science chapter 11 Work and Energy
- NCERT solutions for class 9 science chapter 12 sound
- NCERT solutions for class 9 science chapter 13 Why do We Fall Ill
- NCERT solutions for class 9 science chapter 14 Natural Resources
- NCERT solutions for class 9 science chapter 15 Improvement in Food Resources