{"id":623853,"date":"2023-08-31T12:11:50","date_gmt":"2023-08-31T12:11:50","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=623853"},"modified":"2023-08-31T14:21:56","modified_gmt":"2023-08-31T14:21:56","slug":"623853-2","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/623853-2\/","title":{"rendered":""},"content":{"rendered":"\n\n\n\n\n

EXERCISE 9.1<\/h3>\n\n\n\n

Type 1<\/p>\n\n\n\n

QUESTION 1<\/h4>\n\n\n\n

Fill up the blanks so that the statement becomes true:<\/strong>
(i) If two sides of one triangle and their __ angle are equal to the corresponding two sides and their included angle of the other triangle, then the two triangles are congruent.<\/strong>
(ii) If __ sides of one triangle are respectively equal to the three sides of the other triangle, then the two triangles are congruent.<\/strong>
(iii) If in two triangles ABC and DEF , AB=DF , BC=DE and \u2220B=\u2220D , then \u0394ABC\u2245__<\/strong>
(iv) If in two triangles PQR and DEF , PR=EF , QR=DE and PQ=FD , then \u0394PQR\u2245__<\/strong>
(v) If in a right angled \u0394ABC , M is the mid-point of hypotenuse AC , then BM=\\dfrac{1}{2}<\/strong>__<\/strong>
(vi) In any triangle , sides opposite to equal angles are __<\/strong><\/p>\n\n\n\n

(vii) In any triangle, angles opposite to equal sides are __<\/strong>
(viii) In any \u0394ABC , if AB=BC and \u2220C=80\u00b0 , then \u2220B __<\/strong>
(ix) In any triangle PQR , if \u2220P=\u2220R , then PQ __<\/strong>
(x) In right angled triangles, \u0394PQR and \u0394DEF , if hypotenuse PQ=hypotenuse EF and side PR=side DE , then \u0394PQR\u2245__<\/strong>
(xi) In isosceles \u0394ABC , if AB=AC and BD and CE are altitudes , then BD __CE<\/strong>
(xii) If in \u0394ABC , altitudes CE and BF are equal , then AB=__<\/strong><\/p>\n\n\n\n

\n\n\n\n

QUESTION 2<\/h4>\n\n\n\n

Which of the following statements are true (T) and which are false (F) :<\/strong>
(i) If \u0394ABC\u2245\u0394PQR , then AB=PQ<\/strong>
(ii) If \u0394DEF\u2245\u0394RPQ , then \u2220D=\u2220Q<\/strong>
(iii) If \u0394PQR\u2245\u0394CAB , then PQ=CA<\/strong>
(iv) Each angle of an equilateral triangle is 60\u00b0<\/strong>
(v) Bisectors of equal angles of any triangle are equal.<\/strong>
(vi) The two altitudes corresponding to two equal sides of a triangle need not be equal<\/strong><\/p>\n\n\n\n

(vii) If in \u0394ABC and \u0394DEF , AB=4cm , AC=7cm , \u2220B=45\u00b0 , DE=4cm ,DF=7cm and \u2220D=45\u00b0 , then the two triangles are congruent.<\/strong>
(viii) If in \u0394ABC and \u0394PQR , \u2220A=\u2220P=40\u00b0 , AB=4.5cm , AC=5cm and PQ=4.5cm , QR=5cm , then the two triangles are congruent<\/strong>
(ix) Two equilateral triangles are always congruent.<\/strong>
(x) If in two right angled triangles, one side and one acute angle of a triangle is equal to one side and corresponding angle of the other triangle then two triangle will be congruent.<\/strong><\/p>\n\n\n\n

\n\n\n\n

QUESTION 3<\/h4>\n\n\n\n

(i) Write the Side-Side-Side criterion for congruence of two triangles.<\/strong>
(ii) Write the Side-Angle-Side criterion for congruence of two triangles<\/strong>
Page 9.27
(iii) State the Angle-Side-Angle criterion for congruence of two triangles.<\/strong>
(iv) Write the right angle-hypotenuse-side criterion for congruence of two right angled triangles.<\/strong><\/p>\n\n\n\n

\n\n\n\n

Type 2<\/p>\n\n\n\n

Q4 | Ex-9.1 | Class 9 | Triangles | KC Sinha Solution | myhelper<\/strong><\/p>\n\n\n\n

QUESTION 4<\/h4>\n\n\n\n

In the f\u200cigure below, if l||BC and AB=AC , then find \u2220C.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u2235EF||BC , AB is transversal<\/p>\n\n\n\n

\u2220EAB=\u2220ABC=70\u00b0 (alternate interior angle)<\/p>\n\n\n\n

also, <\/p>\n\n\n\n

In \u0394ABC ,Given : AB=AC thus form isosceles triangle .And angles opposite to equal sides of an isosceles triangle are equal.<\/p>\n\n\n\n

\u2234\u2220B=\u2220C=70\u00b0<\/p>\n\n\n\n

\n\n\n\n

Q5 | Ex-9.1 | Class 9 | Triangles | KC Sinha Solution | myhelper<\/strong><\/h4>\n\n\n\n

QUESTION 5<\/h4>\n\n\n\n

In the f\u200cigure below , if PQ=PR , then find the value of angle \u03b1<\/strong>
\"\"<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

In \u0394PQR , PQ=PR which means its a isosceles triangle. And angles opposite to equal sides of an isosceles triangle are equal.<\/p>\n\n\n\n

\u2234\u2220PQR=\u2220PRQ=60\u00b0<\/p>\n\n\n\n

Also,<\/p>\n\n\n\n

\u2220PRQ+\u2220\u03b1 =180\u00b0 (Linear Pair)<\/p>\n\n\n\n

60\u00b0+\u2220\u03b1 =180\u00b0<\/p>\n\n\n\n

\u2220\u03b1 =180\u00b0-60\u00b0=120\u00b0<\/p>\n\n\n\n

ALTERNATE METHOD<\/strong><\/p>\n\n\n\n

In \u0394PQR , PQ=PR which means its a isosceles triangle. And angles opposite to equal sides of an isosceles triangle are equal.<\/p>\n\n\n\n

\u2234\u2220PQR=\u2220PRQ=60\u00b0<\/p>\n\n\n\n

In \u0394PQR<\/p>\n\n\n\n

\u2220PQR+\u2220PRQ+\u2220QPR=180\u00b0 (Angle sum property of triangle)<\/p>\n\n\n\n

60\u00b0+60\u00b0+\u2220QPR=180\u00b0<\/p>\n\n\n\n

120\u00b0+\u2220QPR=180\u00b0<\/p>\n\n\n\n

\u2220QPR=180\u00b0-120\u00b0=60\u00b0<\/p>\n\n\n\n

Exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles of the triangle<\/p>\n\n\n\n

\u2220\u03b1=\u2220QPR+\u2220PQR<\/p>\n\n\n\n

\u2220\u03b1=60\u00b0+60\u00b0=120\u00b0<\/p>\n\n\n\n

\n\n\n\n

Q6 | Ex-9.1 | Class 9 | Triangles | KC Sinha Solution | myhelper<\/strong><\/h4>\n\n\n\n

QUESTION 6<\/h4>\n\n\n\n

If one side of an equilateral triangle be produced, then what will be the value of the exterior angle ?<\/strong>
Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

In an equilateral triangle all angles are 60\u00b0<\/p>\n\n\n\n

\u2234\u2220ACB=60\u00b0<\/p>\n\n\n\n

\u2220BCA+\u2220ACD=180\u00b0 (linear pair)<\/p>\n\n\n\n

60\u00b0+\u2220ACD=180\u00b0<\/p>\n\n\n\n

\u2220ACD=180\u00b0-60\u00b0<\/p>\n\n\n\n

\u2220ACD=120\u00b0<\/p>\n\n\n\n

So ,exterior angle is 120\u00b0<\/p>\n\n\n\n

\n\n\n\n

Q7 | Ex-9.1 | Class 9 | Triangles | KC Sinha Solution | myhelper<\/strong><\/h4>\n\n\n\n

QUESTION 7<\/h4>\n\n\n\n

(i) In a \u0394ABC , \u2220A=50\u00b0 , \u2220B=80\u00b0 , then what type of triangle it would be ?<\/strong>
(ii) In a \u0394ABC ,AB=AC and \u2220A=100\u00b0 , then write the measure of \u2220B.<\/strong>
Sol :<\/p>\n\n\n\n

(i) <\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

In \u0394ABC <\/p>\n\n\n\n

\u2220A+\u2220B+\u2220C=180\u00b0 (angle sum property of triangle)<\/p>\n\n\n\n

50\u00b0+80\u00b0+\u2220C=180\u00b0<\/p>\n\n\n\n

\u2220C=180\u00b0-130\u00b0=50\u00b0<\/p>\n\n\n\n

As we can see two angles (i.e. \u2220A=\u2220C) are equal . Thus, its a isosceles triangle<\/p>\n\n\n\n

(ii) <\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

In \u0394ABC <\/p>\n\n\n\n

\u2220B=\u2220C which means its a isosceles triangle. And angles opposite to equal sides of an isosceles triangle are equal.<\/p>\n\n\n\n

\u2220A+\u2220B+\u2220C=180\u00b0 (angle sum property of triangle)<\/p>\n\n\n\n

100\u00b0+x+x=180\u00b0<\/p>\n\n\n\n

2x=180\u00b0-100\u00b0<\/p>\n\n\n\n

2x=80\u00b0<\/p>\n\n\n\n

x=40\u00b0<\/p>\n\n\n\n

\n\n\n\n

Q8 | Ex-9.1 | Class 9 | Triangles | KC Sinha Solution | myhelper<\/strong><\/h4>\n\n\n\n

QUESTION 8<\/h4>\n\n\n\n

Write measure of each angle in degrees of an isosceles triangle, if :<\/strong>
(i) Each angle of base is twice that of the vertical angle.<\/strong>
(ii) Each angle of base is four times that of the vertical angle.<\/strong>
(iii) In \u0394ABC  , if \u2220A=100\u00b0 and AB=AC , then find \u2220B and \u2220C<\/strong>
(iv) In an equilateral triangle, write measure of each angle in degrees.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

(i)<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

In \u0394ABC<\/p>\n\n\n\n

\u2220A+\u2220B+\u2220C=180\u00b0 (Angle sum property of triangle)<\/p>\n\n\n\n

x+2x+2x=180\u00b0<\/p>\n\n\n\n

5x=180\u00b0<\/p>\n\n\n\n

x=\\frac{180}{5}<\/p>\n\n\n\n

x=36\u00b0<\/p>\n\n\n\n

2x=2(36\u00b0)=72\u00b0<\/p>\n\n\n\n

Angles are 36\u00b0, 72\u00b0, 72\u00b0<\/p>\n\n\n\n

(ii) <\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

In \u0394ABC<\/p>\n\n\n\n

\u2220A+\u2220B+\u2220C=180\u00b0 (Angle sum property of triangle)<\/p>\n\n\n\n

x+4x+4x=180\u00b0<\/p>\n\n\n\n

9x=180\u00b0<\/p>\n\n\n\n

x=\\frac{180}{9}<\/p>\n\n\n\n

x=20\u00b0<\/p>\n\n\n\n

4x=4(20\u00b0)=80\u00b0<\/p>\n\n\n\n

Angles are 20\u00b0, 80\u00b0, 80\u00b0<\/p>\n\n\n\n

(iii) <\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

In \u0394ABC<\/p>\n\n\n\n

Opposite sides are equal thus \u0394ABC is isosceles , also opposite angle are also equal.<\/p>\n\n\n\n

\u2234\u2220B=\u2220C<\/p>\n\n\n\n

\u2220A+\u2220B+\u2220C=180\u00b0 (Angle sum property of triangle)<\/p>\n\n\n\n

100\u00b0+x+x=180\u00b0<\/p>\n\n\n\n

2x=180\u00b0-100\u00b0<\/p>\n\n\n\n

2x=80\u00b0<\/p>\n\n\n\n

x=40\u00b0<\/p>\n\n\n\n

\u2220B=40\u00b0<\/p>\n\n\n\n

(iv) <\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

In \u0394ABC<\/p>\n\n\n\n

All angles are equal in equilateral triangle.<\/p>\n\n\n\n

x+x+x=180\u00b0 (Angle sum property of triangle)<\/p>\n\n\n\n

3x=180\u00b0<\/p>\n\n\n\n

x=\\frac{180}{3}<\/p>\n\n\n\n

x=60\u00b0<\/p>\n\n\n\n

\n\n\n\n

Q9 | Ex-9.1 | Class 9 | Triangles | KC Sinha Solution | myhelper<\/strong>OPEN IN YOUTUBE<\/strong><\/h4>\n\n\n\n

QUESTION 9<\/h4>\n\n\n\n

(i) In \u0394ABC , AB=AC and side BC is produced to point D , such that \u2220ACD=150\u00b0 , then write the measure of \u2220ABC<\/strong>
(ii) In an isosceles \u0394ABC , AB=AC and BC is produced to point D such that \u2220ACD=116\u00b0 , then write the measure of \u2220BAC<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

(i) <\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

\u2220ACB+\u2220ACD=180\u00b0 (linear pair)<\/p>\n\n\n\n

x+150\u00b0=180\u00b0<\/p>\n\n\n\n

x=180\u00b0-150\u00b0=30\u00b0<\/p>\n\n\n\n

Also,<\/p>\n\n\n\n

Given : AB=AC thus the given \u0394 is isosceles.<\/p>\n\n\n\n

\u2234\u2220ABC=ACB<\/p>\n\n\n\n

\u2220ABC=30\u00b0<\/p>\n\n\n\n

(ii)<\/strong> <\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

\u2220ACB+\u2220ACD=180\u00b0 (linear pair)<\/p>\n\n\n\n

x+116\u00b0=180\u00b0<\/p>\n\n\n\n

x=180\u00b0-116\u00b0=64\u00b0<\/p>\n\n\n\n

Also,<\/p>\n\n\n\n

Given : AB=AC thus the given \u0394 is isosceles.<\/p>\n\n\n\n

\u2234\u2220ABC=ACB<\/p>\n\n\n\n

\u2220ABC=64\u00b0<\/p>\n\n\n\n

Exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles of the triangle<\/p>\n\n\n\n

\u2220ACD=\u2220ABC+\u2220BAC<\/p>\n\n\n\n

116\u00b0=64\u00b0+\u2220BAC<\/p>\n\n\n\n

116\u00b0-64\u00b0=\u2220BAC<\/p>\n\n\n\n

\u2220BAC=52\u00b0<\/p>\n\n\n\n

\n\n\n\n

Q10 | Ex-9.1 | Class 9 | Triangles | KC Sinha Solution | myhelper<\/strong>OPEN IN YOUTUBE<\/strong><\/h4>\n\n\n\n

QUESTION 10<\/h4>\n\n\n\n

(i) In the f\u200cigure below, \u0394ABC is an isosceles triangle, whose sides AB=AC and XY is parallel to BC. If \u2220A=30\u00b0 , then find \u2220BXY in degrees .<\/strong>
<\/a>
Sol :<\/p>\n\n\n\n

=105\u00b0<\/p>\n\n\n\n

\n\n\n\n

Page 9.28<\/p>\n\n\n\n

Q11 | Ex-9.1 | Class 9 | Triangles | KC Sinha Solution | myhelper<\/strong>OPEN IN YOUTUBE<\/strong><\/h4>\n\n\n\n

QUESTION 11<\/h4>\n\n\n\n

In the f\u200cigure below, AP=AQ and BP=BQ , then prove that AB is a bisector of \u2220PAQ and \u2220PBQ.<\/strong>
\"\"
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 12<\/h4>\n\n\n\n

In the f\u200cigure below, diagonal AC of quadrilateral ABCD is bisector of \u2220C and \u2220A . Prove that AB=AD and CB=CD<\/strong>
\"\"
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 13<\/h4>\n\n\n\n

In the f\u200cigure below, \u2220CPD=\u2220BPD and AD is bisector of \u2220BAC. Prove that , \u0394CAP\u2245\u0394BAP and CP=BP<\/strong>
\"\"
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 14<\/h4>\n\n\n\n

In the f\u200cigure below, \u2220BCD=\u2220ADC and \u2220ACB=\u2220BDA. Prove that AD=BC and \u2220A=\u2220B<\/strong>
[Hint: Prove that \u0394ACD\u2245\u0394BDC by A-S-A]
\"\"
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 15<\/h4>\n\n\n\n

In the f\u200cigure below, line segment joining mid points M and N of sides of AB and CD of quadrilateral ABCD is perpendicular to both. Prove that other sides of quadrilateral are equal.<\/strong>
[Hint: Join MD and MC and prove that \u0394MDN\u2245\u0394MCN . Again , prove that \u0394AMD\u2245\u0394BMC]
\"\"
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 16<\/h4>\n\n\n\n

 Prove that \u0394ABC will be an isosceles triangle in which AB=AC , if and only if:<\/strong>
(i) Median AD is perpendicular to BC <\/strong>
(ii) Bisector of \u2220BAC is perpendicular to BC<\/strong>
(iii) Altitude AB , bisects \u2220BAC<\/strong>
(iv) Altitude BE=Altitude CF<\/strong>
(v) Median BE=Median CF<\/strong><\/p>\n\n\n\n

\n\n\n\n

QUESTION 17<\/h4>\n\n\n\n

If bisector of the vertical angle of a triangle bisects the base , then prove that the triangle will be isosceles<\/strong>
\"\"
[Hint : Produce AD such that AD = DE. Join CE (see Fig.)
Proof: In \u0394ABD and \u0394EDC
BD=DC [Given]
AD=DE [construction]
\u2220ADB=\u2220EDC [Vertically opposite angles]
\u0394ABD\u2245\u0394ECD [S-A-S congruence rule]
AB=CE [CPCT]
\u2220BAD=\u2220DEC [CPCT]
\u2220BAD=\u2220DAC [Given]
\u2220DAC=\u2220DEC
AC=CE but AB=CE
AB=AC]
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 18<\/h4>\n\n\n\n

In the f\u200cigure below, RS=QT and QS=RT . Prove that PQ=PR<\/strong>
\"\"
[Hint: By S-S-S , prove that
\u0394RQS\u2245\u0394QRT
\u2234\u2220QRS=\u2220RQT
\u2234PQ=PR
]<\/p>\n\n\n\n

QUESTION 19<\/h4>\n\n\n\n

In the f\u200cigure below, PS=PR , \u2220TPS=\u2220QPR . Prove that PT=PQ.<\/strong>
\"\"
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 20<\/h4>\n\n\n\n

ABC is an isosceles triangle with AB = AC and bisectors of \u2220B and \u2220C intersect at P. Prove that BP=CP and also that AP is bisector of \u2220BAC.<\/strong>
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 21<\/h4>\n\n\n\n

ABC is an isosceles triangle with AB = AC, and AD is perpendicular from A to BC. Prove that AD is median of the triangle.<\/strong>
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 22<\/h4>\n\n\n\n

In the f\u200cigure below, AB=AC and PB = PC , then prove that \u2220ABP=\u2220ACP<\/strong>
\"\"
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 23<\/h4>\n\n\n\n

In the adjoining f\u200cigure , X and Y are points lying on equal sides AD and AC respectively of any \u0394ABC such that AX = AY. Prove that XC = YB.<\/strong>
<fig to be added>
[Hint: AB=AC and AX=AY
\u2234 AB-AX=AC-AY; or XB=YC
Now prove that \u0394XBC\u2245\u0394YCB]
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 24<\/h4>\n\n\n\n

In the f\u200cigure below, AB=AC and BE and CF are respectively bisectors of \u2220B and \u2220C . Prove that BE = CF.<\/strong>
\"\"
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 25<\/h4>\n\n\n\n

In two right angled triangles. one side and an acute angle of one triangle is equal to one side and corresponding acute angle of the other triangle. Prove that the two triangles are congruent.<\/strong>
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 26<\/h4>\n\n\n\n

Altitudes AD and BE of a \u0394ABC are such that AE=BD . Prove that AD=BE<\/strong>
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 27<\/h4>\n\n\n\n

X and Y are two points respectively on the sides AD and BC of the square ABCD such that AY=BX. Prove that BY=AX and \u2220BAY=\u2220ABX<\/strong>
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 28<\/h4>\n\n\n\n

In adjoining f\u200cigure, \u2220QPR=\u2220PQR and M and N are respectively points on QR and PR of \u0394PQR , so that QM=PN . Prove that OP=OQ where O is the points of intersection of PM and QN<\/strong>
<fig to be added>
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 29<\/h4>\n\n\n\n

ABC is a triangle in which \u2220B=2\u2220C and AB=AD. D is a point on the side BC, so that AD bisects \u2220BAC=72\u00b0
[Hint: Let \u2220ACB=x , then \u2220ABC=2x
\u2220ADB=2x [since AB=AD]
Now \u2220ADB=\u2220ACD+\u2220CAD
or 2x=x+\u2220CAD
or \u2220CAD=x
\u2234\u2220DAB=x [since AD bisects \u2220BAC]
In \u0394ABC , x+2x+2x=180\u00b0 or
5x=180\u00b0
x=36\u00b0
\u2220BAC=2x=2\u00d736=72\u00b0
]
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 30<\/h4>\n\n\n\n

Prove that on extending equal sides of an isosceles triangle, bisectors of which are formed below the base form an isosceles triangle.<\/strong>
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 31<\/h4>\n\n\n\n

Prove that , an equiangular triangle is also an equilateral triangle (adjoining fig)<\/strong>
<fig to be added>
[Hint: \u2220A=\u2220B
AC=BC
\u2220B=\u2220C..(i)
AB=AC..(ii)
From (i) and (ii) , AB=BC=AC
]
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 32<\/h4>\n\n\n\n

Line Segments AB and CD intersect each other at O such that <\/strong>
AO=OD and <\/strong>
OB=OC<\/strong>
Prove that AC=BD; but it is not necessary that AC and BD are parallel.<\/strong>
\"\"
[Hint: In \u0394AOC and \u0394DOB
OA=OD, OC=OB
and \u22201=\u22202 [Vertically opposite angles]
\u0394AOC\u2245\u0394DOB [S-A-S congruence rule]
AC=BD
\u22203=\u22205; \u22204=\u22206 [CPCT]
Now \u22203 , \u22205 and \u22204 , \u22206 are neither pair of corresponding angles nor pair of alternate angles. Hence, it is not necessary that AC||BD]
]
Sol :<\/p>\n\n\n\n

\n\n\n\n

QUESTION 33<\/h4>\n\n\n\n

From a point on the bisector of an angle, if a line parallel to any side be drawn meeting other side of the angle, prove that the triangle so formed is an isosceles triangle.<\/strong>
[Hint: Let L be a point on the bisector of \u2220ABC. Front L, a line LM drawn || to AB meets BC in M.]
\"\"
To\u00a0 prove: \u0394LMB is isosceles (Fig. adjoining)
\u22201=\u22202 [\u2235\u00a0 BL is bisector of \u2220ABC]
Also , \u22203=\u22201 [Alternate angles]
\u2234 \u22202=\u22203 or BM=LM
So , \u0394LMB is an isosceles triangle]<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 9 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

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