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MATHS 4 Marks Questions

Triangles · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 22 questions.

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Question 14 Marks
ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE.
Answer
Given $\triangle\text{ABC}$ is an isosceles triangle in which AB = AC and BD, CE are its two medians.
To show BD = CE.In $\triangle\text{ABD}$ and $\triangle\text{ACE},$
$\text{AB}=\text{AC}$
$\angle\text{A}=\angle\text{A}$
$\text{AD}=\text{AC}$
$\frac{1}{2}\text{AB}=\frac{1}{2}\text{AC}$
$\text{AE}=\text{AD}$
AS D is the mid-point of AC and E is the mid-point of AB.
$\triangle\text{ABC}\cong\triangle\text{ACE}$
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Question 24 Marks
ABC is a right triangle with AB = AC. If bisector of meets BC at D,then prove that BC = 2AD.
Answer
In $\triangle\text{ABC}$ is a right angle triangle with AB = AC, is the Bisector.$\text{BC}=2\text{AD}$
$\text{AB}=\text{AC}$
$\angle\text{C}=\angle\text{B}$ 
Now, in right angle $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$
$\Rightarrow 90^{\circ}+\angle\text{B}+\angle\text{B}=180^{\circ}$
$\Rightarrow 2\angle\text{B}=90^{\circ}$
$\Rightarrow \angle\text{B}=45^{\circ}$
$\Rightarrow \angle\text{B}=\angle\text{C}=45^{\circ}$
$\Rightarrow \angle\text{3}=\angle\text{4}=45^{\circ}$
$\Rightarrow \angle\text{1}=\angle\text{2}=45^{\circ}$
$\text{BD}=\text{AD},\text{DC}=\text{AD}$
Hence, $\text{BC}=\text{BD}+\text{CD}$
$=\text{AD}+\text{AD}$
$=2\text{AD}$
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Question 34 Marks
Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than $\frac{2}{3}$ of a right angle.
Answer
Consider $\triangle\text{ABC}$ in which BC is the longest side. To prove $\angle\text{A}=\frac{2}{3}$ right angle Proof In $\triangle\text{ABC},\ \text{BC}>\text{AB}.$
[consider BC is the largest side] $\Rightarrow\ \angle\text{A}>\angle\text{C}$ ...(i) [angle opposite the lngest side is greatest] $\text{and}\ \text{BC}>\text{AC}$  $\Rightarrow\ \angle\text{A}>\angle\text{B},$  [angle opposite the longest side is greatest] On adding Eqs. (i) and (ii), we get $2\angle\text{A}>\angle\text{B}+\angle\text{C}$ $\Rightarrow\ 2\angle\text{A}+\angle\text{A}>\angle\text{A}+\angle\text{C}$ [adding $\angle\text{A}$ both sides] $\Rightarrow\ 3\angle\text{A}>\angle\text{A}+\angle\text{B}+\angle\text{C}$  $\Rightarrow\ 3\angle\text{A}>180^\circ$ [sum of the angles of a triangle is 180°] $\Rightarrow\ \angle\text{A}>\frac{2}{3}\times90^\circ$ i.e., $\angle\text{A}>\frac{2}{3}$ of a right angle Hence proved.
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Question 44 Marks
In D and E are points on side BC of a $\triangle\text{ABC}$ such that BD = CE and AD = AE. Show that $\triangle\text{ABC}\cong\triangle\text{ACE}.$
Answer
Given, $\triangle\text{ABC}$ in which BD = CE and AD = AE
$\triangle\text{ABC}\cong\triangle\text{ACE}$
In $\triangle\text{ADE},$ we have
$\text{AD}=\text{AE}$
 $\angle\text{ADE}=\angle\text{AED}$
Now,
$\angle\text{ADE}+\angle\text{AEC}=180^{\circ}\ ...(\text{i})$
$\angle\text{AED}+\angle\text{AEC}=180^{\circ}\ ...(\text{ii})$
For eq.(i) and (ii),
$\angle\text{ADE}+\angle\text{ADB}=\angle\text{AED}+\angle\text{AEC}$
$\Rightarrow\angle\text{ADB}=\angle\text{AEC}$
In $\triangle\text{ABD},$ we have
$\text{BD}=\text{CE}$
So, by SAS criterion of congrurncr, we have
$\triangle\text{ABC}\cong\triangle\text{ACE}.$
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Question 54 Marks
ABCD is quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.
Answer
Given In quadrilateral ABCD, AB = AD and CB = CD. constrion: To prove AC is the petpendicular bisector of BD.uction Join AC and BD. 
Proof In $\triangle\text{ABC}\text{ and }\triangle\text{ADC},$  $\text{AB}=\text{AD}$ [given] $\text{BC}=\text{CD}$ [given] $\text{and}\ \text{AC}=\text{AC}$ [common side] $\therefore \triangle\text{ABC}\cong\triangle\text{ADC}$ [by SSS congruence rule] $\Rightarrow\ \angle1=\angle2$ [by CPCT] now, in $\triangle\text{AOB}\text{ and }\triangle\text{AOD},\ \text{AB}=\text{AD}$ [given] $\Rightarrow\ \angle1=\angle2$ [proved above] $\text{and}\ \text{AO}=\text{AO} $ [common side] $\therefore\ \triangle\text{AOB}\cong\triangle\text{AOD}$ [by SAS congruence rule] $\Rightarrow\ \text{BO}=\text{DO}$ [by CPCT] $\text{and}\ \angle3=\angle4$ [by CPCT]...(i) But $\angle3+\angle4=180^\circ$[linear pair axiom] $\angle3+\angle3=180^\circ$ [from Eq. (i)] $\Rightarrow\ 2\angle3=180^\circ$ $\Rightarrow\ \angle3=\frac{180^\circ}{2}$ $\therefore\ \angle3=90^\circ$ i.e., AC is perpendicular bisector of BD.
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Question 64 Marks
In a triangle ABC, D is the mid-point of side AC such that $\text{BD}=\frac{1}{2}\text{AC}.$Show that $\angle\text{ABC}$ is a right angle. 
Answer
We have to prove that $\angle\text{ABC}=90^\circ$ As is the mid-point of AC, so, AD = DC  Also, $\text{BD}=\frac{\text{1}}{\text{2}}\text{AC}=\text{AD}$  $\big[\therefore$ D is the mid-point of AC$\big]$ $\therefore$  BD = AD = DC
In $\triangle\text{ABC},$ We have BD = AD $\therefore\angle1=\angle2\ ...(1)$ $\big[\because$ Angles opposite to equal sides are equal$\big]$ In $\triangle\text{BCD},$We have BD = DC, $\therefore\angle\text{3}=\angle4 ...(2)$ In $\triangle\text{ABC},$ We have $\angle1+\angle\text{ABC}+\angle4=180^\circ$ $\Rightarrow\angle\text{1}+\angle\text{2}+\angle\text{3}\angle+\text{4}=180^\circ$ $\Rightarrow2(\angle\text{2}+\angle\text{3})=180^\circ)$ $\Rightarrow\angle2+\angle3=90^\circ$ $\Rightarrow\angle=90^\circ$ Hence proved.
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Question 74 Marks
S is any point on side QR of a $\triangle\text{PQR}.$ show that PQ + QR + RP > 2PS.
Answer
Given A point S on QR.In $\triangle\text{PQS},$ we have
$\text{PQ}+\text{QS}>\text{PS}\ ...(\text{i})$
Now, in $\triangle\text{PSR},$ we have
$\text{RS}+\text{RP}>\text{PS}\ ...(\text{ii})$
From eq.(i) and (ii),
$\text{PQ}+\text{QS}+\text{RS}+\text{RP}>2\text{PS}$
$\Rightarrow\text{PQ}+\text{QR}+\text{RP}>2\text{PS}$
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Question 84 Marks
ABC is an isoscles triangle in which AC = BC. AD are respectively two altitudes to sides BC and BC. prive that AE = BD.
Answer
In $\triangle\text{ABC}$ and $\angle\text{DBC}$ we have $\text{AB}=\text{AC}$ $\angle\text{ABD}=\angle\text{ACD}$ $\angle\text{BAD}=\angle\text{CAD}$ $\therefore\triangle\text{ADC}\cong\triangle\text{BCD}$  $\therefore\text{AB}=\text{AC}$
$\text{AC}-\text{CE}=\text{BC}-\text{CD}$ $\Rightarrow\text{AE}=\text{BD}$
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Question 94 Marks
Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.
Answer
Givan in $\triangle\text{ABC},$ AD is a median. 
Construction produce AD to a point E such that AD = DE and join CE. To prove AC + AB > 2AD Proof in $\triangle\text{ABD}\text{ and } \triangle \text{ECD},$ $\text{AD}=\text{DE}$ [by construction]  $\text{BD}=\text{CD}$ [given AD is the media] and $\angle\text{ADB}=\angle\text{CDE}$  [vertically opposite angle] $\therefore\ \triangle\text{ABD}\cong\text{ECD}$ [by SAS congruence rule]  $\Rightarrow\ \text{AB}=\text{CE}$ [by CPCT]...(i) Now, in $\triangle\text{AEC},$ $\text{AC}+\text{EC}>\text{AE}$  $\therefore\ \text{AC}+\text{AB}>2\text{AD} $ [Sum of two sides of a triangle is greater than the third side] [from Eq. (i) and also taken that AD = DE] Hence proved.
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Question 104 Marks
ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.
Answer
Given two $\triangle\text{ABC}$ and $\angle\text{DBC}$ are formed on the same base BC Such that A and D lie on the opposite side of AB = AC and DB = DC.
In $\triangle\text{ABD}$ and $\triangle\text{ACD},$
$\text{AB}=\text{AC}$
$\angle\text{ABD}=\angle\text{ACD}$
$\angle\text{BAD}=\angle\text{CAD}$In $\triangle\text{ABO},$
$\text{AB}=\text{AC}$
$\angle\text{AOB}+\angle\text{AOB}=180^{\circ}$
$2\angle\text{AOB}=180^{\circ}$
$\angle\text{AOB}=\frac{180^{\circ}}{2}=180^{\circ}$
Hence, AD is the perpendicular of BC.
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Question 114 Marks
Q is point on the side SR of a $\triangle\text{PSR}$ such that PQ = PR. prove that PS > PQ.
Answer
Given, PQ = PRIn $\triangle\text{PRQ},$ we have
$\text{PR} =\text{PQ}$
$\Rightarrow\angle\text{1}=\angle\text{R}$
But, $\angle\text{1}>\angle\text{S}$
$\Rightarrow\angle\text{R}>\angle\text{S}$
$\Rightarrow\text{PS}>\text{PR}$
$\Rightarrow\text{PS}=\text{PQ}$
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Question 124 Marks
ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.
Answer
Given in right angled $\triangle\text{ABC},\text{AB}= \text{AC}\text{ and }\text{CD}$ is the bisector of $\angle\text{C}.$ contruction draw $\text{DE}\ \bot\ \text{BC}.$ to prove $\text{AC}+\text{AD}= \text{BC}$ proof in right angled $\triangle\text{ABC},\text{AB}=\text{AC}\text{ and }\text{BC}$ is a hypotenuse [given]
$\angle1=\angle2$ [given,$\text{CD}$ isthe bisector of $\angle\text{C}$] $\text{DC}=\text{DC}$ [common sides] $\therefore\ \triangle\text{DAC}\cong\triangle\text{DEC}$ [by AAS congruence rule] $\Rightarrow\ \text{DA}=\text{DE}.$ [by CPCT]...(i) $\text{and}\ \text{AC}=\text{EC}$ ...(ii) $\text{ln}\ \triangle\text{ABC},\ \text{AB}=\text{AC}$ $\angle\text{C}=\angle{B}$   [angle opposite to equal sides are equal]...(iii) Again,in $\triangle\text{ABC},\ \angle\text{A}+\angle\text{B}+\angle\text{C}= 180^\circ$ [by angle sum property of a triangle]  $\Rightarrow\ 90^\circ+\angle\text{B}+\angle\text{B}=180^\circ$ $\Rightarrow\ 2\angle\text{B}=180^\circ-90^\circ$[from Eq.(iii)] $\Rightarrow\ 2\angle\text{B}=90^\circ$ $\Rightarrow\ \angle\text{B}=45^\circ$ $\text{ln}\ \triangle\text{BED},\ \angle5=180^\circ-\big(\angle\text{B}+\angle{4}\big)$  [by angle sum property of a triangle] $=180^\circ-(45^\circ+90^\circ)$ $=180^\circ-135^\circ=45^\circ$ $\therefore\ \angle\text{B}=\angle{5}$ $\Rightarrow\ \text{DE}=\text{BE}\ [\because$ sides opposite to equal angle are equal$]$ ...(iv) from Eqs. (i) and (iv), $\text{DA}=\text{DE}=\text{BE}$ ...(v) $\because\text{BC}=\text{CE}+\text{EB}$  $=\text{CA}+\text{DA}$ [from Eqs. (ii) and (v)] $\therefore\text{AD}+\text{AC}=\text{BC}$ Hence proved.
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Question 134 Marks
ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.
Answer
Given In quadrilateral ABCD, AB is the smallest and CD is largest side to find $\angle\text{B}>\angle\text{D}\text{ or }\angle\text{D}>\angle\text{B}.$
Construction Join BD. Now, in $\triangle\text{ABD},\ \text{AD}>\text{AB}$ [since, AB is the smallest side in ABCD] $\Rightarrow\ \angle{1}>\angle{3}$ [angle opposite to larger side is greater]...(i)  $\text{In }\triangle\text{ABCD},\ \text{CD}>\text{BC}$ [since, CD is the largest side in ABCD] $\Rightarrow\ \angle{2}>\angle{4}$ [angle opposite to larger side is greater]...(ii) On adding Eqs.(i) and (ii), we get $\angle1+\angle2+\angle3+\angle4$  Hence, $\angle\text{B}>\angle\text{D}$
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Question 144 Marks
The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown in figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.
Answer
Given, An object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown in figure.
The image is as far behind the mirror as the object is in front of the mirror OB = OA.$\therefore\text{CN}\perp\text{LM}$ and $\text{AB}\perp\text{LM}$
$\text{AB}\ ||\ \text{CN}$
$\angle\text{A}=\angle\text{i}\ ...(\text{i})$
$\angle\text{B}=\angle\text{r}\ ...(\text{ii})$
$\angle\text{i}=\angle\text{r}\ ...(\text{iii})$
From eq.(i), (ii) and (iii)
$\angle\text{A}=\angle\text{B}$
 In $\triangle\text{COB},$
$\angle\text{B}=\angle\text{A}$
$\angle\text{1}=\angle\text{2}$
$\text{OB}=\text{OA}$
In $\triangle\text{OBC},$
$\angle\text{1}=\angle\text{2}$
$\angle\text{i}=\angle\text{r}$
On mulitiplying both sides of then adding both sides,
$90^{\circ}-\angle\text{i}=90^{\circ}-\angle\text{r}$
$\Rightarrow \angle\text{ACO}=\angle\text{BCO}$
$\Rightarrow \text{OC}=\text{OC}$
$\therefore\triangle\text{OBC}=\triangle\text{OAC}$
$\Rightarrow \text{OB}=\text{OA}$
Hence, the image is as far behind the mirror as the object is in front of the mirror.
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Question 154 Marks
Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. BO is produced to a point M. Prove that $\angle\text{MOC}=\angle\text{ABC}.$
Answer
Given Lines, OB and OC are angle bisectore of $\angle\text{B}$ and $\angle\text{C}$ an isosceles $\triangle\text{ABC}$ such that AB = AC which intersrct each other at O and BO is produced to M.
$\angle\text{MOC}=\angle\text{ABC}$In $\triangle\text{ABC},$
$\text{AB}=\text{AC}$
$\frac{1}{2}\angle\text{ACB}=\frac{1}{2}\angle\text{ABC}$
$\angle\text{OCB}=\angle\text{OBC}$
Now,
$\angle\text{MOC}=\angle\text{OBC}+\angle\text{OCB}$
$\Rightarrow\angle\text{MOC}=\angle\text{OBC}+\angle\text{OBC}$
$\Rightarrow \angle\text{MOC}=2\angle\text{OBC}$
$\Rightarrow \angle\text{MOC}=\angle\text{ABC}$
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Question 164 Marks
Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD
Answer
Given ABCD is a quadilateral,
Construction join diagonais AC and BD. To show AB + BC + CD + DA > AC + BD  In $\triangle\text{ABC},$ AB + BC > AC [Sum of two sides of a triangle is greater than the third side] ...(i) In $\triangle\text{BCD},$ BC + CD > BD [Sum of two sides of a triangle is greater than the third side] ...(ii) In $\triangle\text{CDA},$ CD + DA > AC [Sum of two sides of a triangle is greater than the third side] ...(iii) In $\triangle\text{DAB},$ DA + AB + > BD [Sum of two sides of a triangle is greater than the third side] ...(iv) On adding Eqs. (i), (ii), (iii) (iv), We get 2(AB + BC + CD + DA) > 2 (AC + BD) ⇒ AB + BC + CD + DA > AC + BD
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Question 174 Marks
In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.
Answer
We have to prove that $\angle\text{ABC}=90^\circ$ As is the mid-point of AC, so, AD = DC  Also, $\text{BD}=\frac{\text{1}}{\text{2}}\text{AC}=\text{AD}$  $\big[\therefore$ D is the mid-point of AC$\big]$ $\therefore$  BD = AD = DC
In $\triangle\text{ABC},$ We have BD = AD $\therefore\angle1=\angle2\ ...(1)$ $\big[\because$ Angles opposite to equal sides are equal$\big]$ In $\triangle\text{BCD},$We have BD = DC, $\therefore\angle\text{3}=\angle4 ...(2)$ In $\triangle\text{ABC},$ We have $\angle1+\angle\text{ABC}+\angle4=180^\circ$ $\Rightarrow\angle\text{1}+\angle\text{2}+\angle\text{3}\angle+\text{4}=180^\circ$ $\Rightarrow2(\angle\text{2}+\angle\text{3})=180^\circ)$ $\Rightarrow\angle2+\angle3=90^\circ$ $\Rightarrow\angle=90^\circ$ Hence proved. 
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Question 184 Marks
O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle.
Answer
Given O is a point in the interior of a squaren ABCD such that an euailateral triangle.Since, AOB is an triangle.
$\angle\text{OAB}=\angle\text{OBA}=60^{\circ}\ ...(\text{i})$
$\angle\text{DAB}=\angle\text{CBA}=90^{\circ}\ ...(\text{ii})$
From eq. (i) and (ii)
$\angle\text{DAB}-\angle\text{OAB}=\angle\text{CBA}=\angle\text{OBA}$
$\angle\text{DAO}=\angle\text{CBO}=90^{\circ}$
$\Rightarrow\text{AO}=\text{BO}$
$\Rightarrow \text{AD}=\text{BC}$
$\Rightarrow \text{OC}=\text{OD}$
Hence, an isosceles triangle.
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Question 194 Marks
Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)
Answer
Given, A quadrilateral ABCD.
To prove: AB + BC + CD + DA  < 2(BD + AC) Proof: In $\triangle\text{AOD}$ We have $\therefore\text{OA}+\text{OB}>\text{AB }...(\text{i}) $ $\big[\therefore$ Sum of the lengths of any sides of a triangle must be greater than rhird side$\big]$  In $\triangle\text{BOC}$ We have OB + OC + > BC ...(2) [Same reason] In $\triangle\text{COD},$ We have OC + OD + > CD ...(3) [Same reason] In $\triangle\text{DOA},$ We have OD + OA > DA ...(4)  [Same reason] Adding (1), (2), (3) and (4), We get OA + B + OB + OC + OC + OD + OA + > AB + BC + CD +DA  ⇒ 2(OA + OB + OC + OD) > AB + BC + CD + DA ⇒ 2{(OA + OC) + (OB + OD)} > AB + BC + CD + DA ⇒ 2(AC + BD) > AB + BC + CD + DA ⇒ AB + BC + CD + DA < 2(BD + AC) Hence, proved.
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Question 204 Marks
P is point on the bisector of $\angle\text{ABC}.$ if the line throught P, parallel to BA meet BC at Q, prove thatBPQ is an isosceles triangle.
Answer
We have to prove thet BPQ is an isosceles triangle.$\angle\text{1}=\angle\text{2}\ ...(\text{i})$
Now, PQ is paraller to BA and BP cuts them
$\therefore \angle\text{1}=\angle\text{3}\ ...(\text{ii})$
from (i) and (ii),
In $\triangle\text{BPQ},$ we have
$\angle\text{2}=\angle\text{3}$
$\therefore\text{PQ}=\text{BQ}$
Hence, BPQ is an isosceles triangle.
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Question 214 Marks
Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC Prove that AD = BC.
Answer
Given in trapezium ABCD, points M and N are the mid-points of parallel sides AB and DC respectively and join MN, which is perpendicular to AB and DC. To prove: proof since, M is the mid-point of AB $\therefore \text{AM} = \text{MB}$  Now, in $\triangle \text{AMN} \text{ and } \triangle \text{BMN},\ \text{AM} = \text{MB}$ [proved above] $\angle 3 = \angle 4$ [each 90°] $\text{MN} = \text{MN}$ [common side] $\therefore \triangle \text{AMN} \cong \text{BMN}$ [by SAS congruence rule] $\therefore \angle 1 = \angle 2$ [by CPCT] On multiplying both sides of above equation by - 1 and than adding 90° both sides, we get $90^\circ - \angle 1 = 90^\circ - \angle 2$ $\Rightarrow \angle \text{AND} = \angle \text{BNC}$
Now, in $\triangle \text{ADN} \text{ and } \triangle \text{BNC}$ $\angle \text{AND} = \angle \text{BNC}$ [from Eq.(i)] $\text{AN} = \text{BN}$ $\big[\because\ \triangle\text{AMN}\ \cong\ \triangle\text{BMN}\big]$ $\text{and}\ \text{DN}\ =\ \text{NC}$ $\big[\because $ N is the mid-point of CD (given)$\big]$ $\therefore\ \triangle\text{ADN}\ \cong\ \triangle\text{BCN}$ [by SAS congruence rule] Hence,$\text{AD}\ =\ \text{BC}$ [by CPCT] Hence proved.
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Question 224 Marks
Question 15: Two lines l and m intersect at the point 0 and P is a point on a line n passing through the point 0 such that P is equidistant from l and m. Prove that n is the bisector of the angle formed by l and m.
Answer
Giva two lines L and M intersect at the point O and P is a point on a line n passing through O such that P is equidistant from L and M, i.e., PQ = PR,

To prove N is the bisector of the angle formed by L and M i.e., n is the bisector of $\angle\text{QOR},$
Proof In $\triangle\text{OQP}\text{ and }\triangle\text{ORP} $,
$\angle\text{PQO}=\angle\text{PRO}=\text{90}^0$
[since,P in equidistant from L and M, so PQ and PR should be perpendicular to lines L and M respectively]
OP = OP [commo side]
P = PR [given]
$\therefore\triangle\text{OQP}\cong\triangle\text{ORP}$ [by RHS congruence rule]
$\Rightarrow\angle\text{POQ}=\angle\text{POR}$ [by CPCT]
Hence,n is the bisector of $\angle\text{QOR}.$Hence proved.
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