Question 1

Q1: In quadrilateral ACBD, $\mathrm{AC}=\mathrm{AD}$ and AB bisects $\angle \mathrm{A}$. Show that $	riangle \mathrm{ABC} \cong 	riangle \mathrm{ABD}$....

Accepted Answer

Given: In quadrilateral ACBD: 1. $\mathrm{AC} = \mathrm{AD}$ 2. AB bisects $\angle \mathrm{A}$, which means $\angle \mathrm{CAB} = \angle \mathrm{DAB}$. To Prove: 1. $	riangle \mathrm{ABC} \cong 	riangle \mathrm{ABD}$ 2. State the relationship between BC and BD. Proof: Consider $	riangle \mathrm{...

Question 2

Q2: ABCD is a quadrilateral in which $\mathrm{AD}=\mathrm{BC}$ and $\angle \mathrm{DAB}=\angle \mathrm{CBA}$. Prove that (i) $	riangle \mathrm{ABD} \cong...

Accepted Answer

Given: In quadrilateral ABCD: 1. $\mathrm{AD} = \mathrm{BC}$ 2. $\angle \mathrm{DAB} = \angle \mathrm{CBA}$ To Prove: (i) $	riangle \mathrm{ABD} \cong 	riangle \mathrm{BAC}$ (ii) $\mathrm{BD} = \mathrm{AC}$ (iii) $\angle \mathrm{ABD} = \angle \mathrm{BAC}$ Proof: (i) Proving $	riangle \mathrm{ABD...

Question 3

Q3: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Accepted Answer

Given: 1. AD and BC are perpendicular to AB. This means $\angle \mathrm{OAD} = 90^\circ$ and $\angle \mathrm{OBC} = 90^\circ$. 2. $\mathrm{AD} = \mathrm{BC}$. 3. CD intersects AB at point O. To Prove: CD bisects AB, which means we need to show that $\mathrm{OA} = \mathrm{OB}$. Proof: Consider $	ria...

Question 4

Q4: $l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$. Show that $	riangle \mathrm{ABC} \cong 	riangle \mathr...

Accepted Answer

Given: 1. Line $l \parallel$ line $m$. 2. Line $p \parallel$ line $q$. 3. These lines intersect to form a quadrilateral ABCD, where AB is on line $p$, CD is on line $p$, BC is on line $m$, and DA is on line $l$. This implies ABCD is a parallelogram. To Prove: $$	riangle \mathrm{ABC} \cong 	riangle...

Question 5

Q5: Line $l$ is the bisector of an angle $\angle \mathrm{A}$ and B is any point on $l$. BP and BQ are perpendiculars from B to the arms of $\angle \mathrm...

Accepted Answer

Given: 1. Line $l$ bisects $\angle \mathrm{A}$. This means $\angle \mathrm{PAB} = \angle \mathrm{QAB}$. 2. B is a point on line $l$. 3. $\mathrm{BP} \perp \mathrm{AP}$ (arm of $\angle \mathrm{A}$), so $\angle \mathrm{BPA} = 90^\circ$. 4. $\mathrm{BQ} \perp \mathrm{AQ}$ (arm of $\angle \mathrm{A}$),...

Question 6

Q6: In the given figure, $\mathrm{AC}=\mathrm{AE}, \mathrm{AB}=\mathrm{AD}$ and $\angle \mathrm{BAD}=\angle \mathrm{EAC}$. Show that $\mathrm{BC}=\mathrm{...

Accepted Answer

Given: In the figure formed by triangles ABC and ADE: 1. $\mathrm{AC} = \mathrm{AE}$ 2. $\mathrm{AB} = \mathrm{AD}$ 3. $\angle \mathrm{BAD} = \angle \mathrm{EAC}$ To Prove: $\mathrm{BC} = \mathrm{DE}$ Proof: We are given that $\angle \mathrm{BAD} = \angle \mathrm{EAC}$. Let us add $\angle \mathrm{DA...

Question 7

Q7: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that $\angle \mathrm{BAD}=\angle \mathrm{ABE}$ and $\angle...

Accepted Answer

Given: 1. P is the mid-point of line segment AB, so $\mathrm{AP} = \mathrm{BP}$. 2. $\angle \mathrm{BAD} = \angle \mathrm{ABE}$ (which can be written as $\angle \mathrm{DAP} = \angle \mathrm{EBP}$). 3. $\angle \mathrm{EPA} = \angle \mathrm{DPB}$. To Prove: (i) $	riangle \mathrm{DAP} \cong 	riangle...

Question 8

Q8: In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that $\mathrm{DM}=\math...

Accepted Answer

Given: 1. $	riangle \mathrm{ABC}$ is a right-angled triangle with $\angle \mathrm{C} = 90^\circ$. 2. M is the mid-point of the hypotenuse AB, so $\mathrm{AM} = \mathrm{BM}$. 3. C, M, D are collinear and $\mathrm{DM} = \mathrm{CM}$. To Prove: (i) $	riangle \mathrm{AMC} \cong 	riangle \mathrm{BMD}$...

Question 9

Q1: In an isosceles triangle ABC, with $\mathrm{AB}=\mathrm{AC}$, the bisectors of $\angle \mathrm{B}$ and $\angle \mathrm{C}$ intersect each other at O....

Accepted Answer

Given: 1. $	riangle \mathrm{ABC}$ is an isosceles triangle with $\mathrm{AB} = \mathrm{AC}$. 2. BO bisects $\angle \mathrm{B}$ and CO bisects $\angle \mathrm{C}$. 3. The bisectors intersect at O. To Prove: (i) $\mathrm{OB} = \mathrm{OC}$ (ii) AO bisects $\angle \mathrm{A}$ Proof: (i) Proving $\math...

Question 10

Q2: In $	riangle \mathrm{ABC}, \mathrm{AD}$ is the perpendicular bisector of BC. Show that $	riangle \mathrm{ABC}$ is an isosceles triangle in which $\m...

Accepted Answer

Given: In $	riangle \mathrm{ABC}$, AD is the perpendicular bisector of BC. This means: 1. AD is perpendicular to BC, so $\angle \mathrm{ADB} = \angle \mathrm{ADC} = 90^\circ$. 2. AD bisects BC, so $\mathrm{BD} = \mathrm{CD}$. To Prove: $	riangle \mathrm{ABC}$ is an isosceles triangle with $\mathrm...

Question 11

Q3: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.

Accepted Answer

Given: 1. $	riangle \mathrm{ABC}$ is an isosceles triangle with $\mathrm{AB} = \mathrm{AC}$. 2. BE is an altitude to side AC, so $\mathrm{BE} \perp \mathrm{AC}$ and $\angle \mathrm{BEA} = 90^\circ$. 3. CF is an altitude to side AB, so $\mathrm{CF} \perp \mathrm{AB}$ and $\angle \mathrm{CFA} = 90^\c...

Question 12

Q4: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that (i) $	riangle \mathrm{ABE} \cong 	riangle \mathrm{ACF}$ (ii)...

Accepted Answer

Given: 1. In $	riangle \mathrm{ABC}$, BE is the altitude to AC ($\angle \mathrm{AEB} = 90^\circ$) and CF is the altitude to AB ($\angle \mathrm{AFC} = 90^\circ$). 2. The altitudes are equal, $\mathrm{BE} = \mathrm{CF}$. To Prove: (i) $	riangle \mathrm{ABE} \cong 	riangle \mathrm{ACF}$ (ii) $\ma...

Question 13

Q5: ABC and DBC are two isosceles triangles on the same base BC. Show that $\angle \mathrm{ABD}=\angle \mathrm{ACD}$.

Accepted Answer

Given: 1. $	riangle \mathrm{ABC}$ is an isosceles triangle on base BC, so $\mathrm{AB} = \mathrm{AC}$. 2. $	riangle \mathrm{DBC}$ is an isosceles triangle on base BC, so $\mathrm{DB} = \mathrm{DC}$. 3. Vertices A and D are on the same side of BC. To Prove: $\angle \mathrm{ABD} = \angle \mathrm{ACD...

Question 14

Q8: Show that the angles of an equilateral triangle are $60^{\circ}$ each.

Accepted Answer

Given: $	riangle \mathrm{ABC}$ is an equilateral triangle. This means all its sides are equal: $\mathrm{AB} = \mathrm{BC} = \mathrm{CA}$. To Prove: $\angle \mathrm{A} = \angle \mathrm{B} = \angle \mathrm{C} = 60^\circ$. Proof: In $	riangle \mathrm{ABC}$, we have $\mathrm{AB} = \mathrm{AC}$. Since...

Question 15

Q1: $	riangle \mathrm{ABC}$ and $	riangle \mathrm{DBC}$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC....

Accepted Answer

Given: 1. $	riangle \mathrm{ABC}$ is isosceles with $\mathrm{AB} = \mathrm{AC}$. 2. $	riangle \mathrm{DBC}$ is isosceles with $\mathrm{DB} = \mathrm{DC}$. 3. AD is extended to meet BC at P. To Prove: (i) $	riangle \mathrm{ABD} \cong 	riangle \mathrm{ACD}$ (ii) $	riangle \mathrm{ABP} \cong 	ria...

Question 16

Q2: AD is an altitude of an isosceles triangle ABC in which $\mathrm{AB}=\mathrm{AC}$. Show that (i) AD bisects BC (ii) AD bisects $\angle \mathrm{A}$.

Accepted Answer

Given: 1. $	riangle \mathrm{ABC}$ is an isosceles triangle with $\mathrm{AB} = \mathrm{AC}$. 2. AD is an altitude to BC, so $\mathrm{AD} \perp \mathrm{BC}$ which means $\angle \mathrm{ADB} = \angle \mathrm{ADC} = 90^\circ$. To Prove: (i) AD bisects BC (i.e., $\mathrm{BD} = \mathrm{CD}$) (ii) AD bis...

Question 17

Q3: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $	riangle \mathrm{PQR}$. Show that:...

Accepted Answer

Given: In $	riangle \mathrm{ABC}$ and $	riangle \mathrm{PQR}$: 1. $\mathrm{AB} = \mathrm{PQ}$ 2. $\mathrm{BC} = \mathrm{QR}$ 3. Median $\mathrm{AM} = $ Median $\mathrm{PN}$ To Prove: (i) $	riangle \mathrm{ABM} \cong 	riangle \mathrm{PQN}$ (ii) $	riangle \mathrm{ABC} \cong 	riangle \mathrm{PQR}...

Question 18

Q4: BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Accepted Answer

Given: 1. In $	riangle \mathrm{ABC}$, BE is an altitude to AC, so $\angle \mathrm{BEC} = 90^\circ$. 2. CF is an altitude to AB, so $\angle \mathrm{CFB} = 90^\circ$. 3. The altitudes are equal, $\mathrm{BE} = \mathrm{CF}$. To Prove: $	riangle \mathrm{ABC}$ is isosceles (i.e., $\mathrm{AB} = \mathrm...

Question 19

Q5: ABC is an isosceles triangle with $\mathrm{AB}=\mathrm{AC}$. Draw $\mathrm{AP} \perp \mathrm{BC}$ to show that $\angle \mathrm{B}=\angle \mathrm{C}$.

Accepted Answer

Given: 1. $	riangle \mathrm{ABC}$ is an isosceles triangle with $\mathrm{AB} = \mathrm{AC}$. 2. A perpendicular AP is drawn from A to BC, so $\mathrm{AP} \perp \mathrm{BC}$. This means $\angle \mathrm{APB} = \angle \mathrm{APC} = 90^\circ$. To Prove: $\angle \mathrm{B} = \angle \mathrm{C}$ Proof: C...

NCERT Solutions