Premium Online Home Tutors

3 Tutor System

Starting just at 265/hour

(i) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\)

(ii)\(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\)

(iii) AP bisects \(\angle{A}\) as well as \(\angle{D}\)

(iv) AP is the perpendicular bisector of BC.

Answer :

Given: \(\triangle{ABC}\) and \(\triangle{DBC}\) are two isosceles triangles having same base BC, such that AB = AC and BD = CD.

Proof:

In \(\triangle{ABD}\) and \(\triangle{ACD}\), we have,

AB = AC ...(Given)

BD = CD ...(Given)

and AD is Common side.

Therefore, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\) ...(By SSS Congruency test)

Hence, part (i) is proved.

In \(\triangle{ABP}\) and \(\triangle{ACP}\), we have,

AB = AC ...(Given)

Also, \(\angle{a}\) = \(\angle{b}\) ...(\(\because \) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\))

and AP is the common side.

Therefore,
\(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\) ...(By SAS congruency test)

Hence, part (ii) is proved.

\(\angle{a}\) = \(\angle{b}\) ...(\(\because \) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\))

Hence, proved that AP bisects \(\angle{A}\)

Now, \(\angle{ADB}\) = \(\angle{ACD}\) ...(\(\because \)\(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\))

Now, by subtracting from \(180^\circ\), we get,

\(180^\circ\) - \(\angle{ADB}\) = \(180^\circ\) - \(\angle{ACD}\)

\(\Rightarrow \) \(\angle{BDP}\) = \(\angle{ACD}\) ...(By linear pair axiom)

Hence, proved that AP bisects \(\angle{D}\)

Hence, part (iii) is proved.

Now, BP = CP (Since, \(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\))

and \(\angle{c}\) = \(\angle{d}\) ...(i)

But \(\angle{c}\) + \(\angle{d}\) = \(180^\circ\) ...(Linear pair axiom)

\(\Rightarrow \)\(\angle{c}\) + \(\angle{c}\) = \(180^\circ\) ...(from (i))

\(\Rightarrow \) 2 \(\angle{c}\) = \(180^\circ\)

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

\(\Rightarrow \)\(\angle{c}\) = \(\angle{d}\) = \(90^\circ\)

Therefore, AP is the perpendicular bisector of BC.

Hence, proved.

- AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects \(\angle{A}\)
- The two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \(\triangle{PQR}\) (see figure). Show that (i) \(\triangle{ABM}\) \(\displaystyle \cong \) \(\triangle{PQN}\)(ii) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{PQR}\)
- BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
- ABC is an isosceles triangle with AB = AC. Draw AP perpendicular to BC to show that \(\angle{B}\) = \(\angle{C}\)

- 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