In the figure alongside,
AB = AC
∠A = 48° and
∠ACD = 18°
Show that BC = CD.
In ΔABC,
∠BAC + ∠ACB + ∠ABC = 180°
48° + ∠ACB + ∠ABC = 180°
But ∠ACB = ∠ABC ........[ AB = AC ]
2∠ABC = 180° - 48°
2∠ABC = 132°
∠ABC = 66° = ∠ACB ..........(i)
∠ACB = 66°
∠ACD + ∠DCB = 66°
18° + ∠DCB = 66°
∠DCB = 48° ..........(ii)
Now, In ΔDCB,
∠DBC = 66° .......[ From (i), Since ∠ABC = ∠DBC ]
∠DCB = 48° .......[From (ii)]
∠BDC = 180°- 48° - 66°
∠BDC = 66°
Since ∠BDC = ∠DBC
Therefore, BC = CD
Equal angles have equal sides opposite to them.
Calculate:
(i) ∠ADC
(ii) ∠ABC
(iii) ∠BAC
Given: ACE = 130°; AD = BD = CD
Proof:
(i) ∠ACD + ∠ACE = 18° ....... [ DCE is a st. line ]
⇒ ∠ACD = 180° − 130°
⇒ ∠ACD = 50°
Now, CD = AD
⇒ ∠ACD = ∠DAC = 50° ..... (i)[ Since angels opposite to equal sides are equal]
In ΔADC,
∠ACD = ∠DAC = 50°
∠ACD + ∠DAC + ∠ADC = 180°
50° + 50° + ∠ADC = 180°
∠ADC =180° − 100°
∠ADC = 80°
(ii) ∠ADC = ∠ABD + ∠DAB ....[Exterior angle is equal to sum of opp. interor angle]
But AD = BD
∴ ∠DAB = ∠ABD
⇒ 80° = ∠ABD + ∠ABD
⇒ 2∠BD = 80°
⇒ ∠ABD = 40° = ∠DAB .....(ii)
(iii) ∠BAC = ∠DAB + ∠DAC
substituting the value from (i) and (ii)
∠BAC = 40° + 50°
⇒ ∠BAC = 90°
In the following figure, AB = AC; BC = CD and DE are parallel to BC.
Calculate:
(i) ∠CDE
(ii) ∠DCE
∠FAB = 128° .......[ Given ]
∠BAC + ∠FAB = 180° ......[ FAC is a st. line ]
⇒ ∠BAC = 180° − 128°
⇒ ∠BAC = 52°
In ΔABC,
∠A = 52°
∠B= ∠C .....[Given AB = AC and angels opposite to equal sides are equal]
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠B + ∠B = 180°
⇒ 52° + 2∠B = 180°
⇒ 2∠B = 128°
⇒ ∠B = 64° = ∠C ........(i)
⇒ ∠B = ∠ADE .......[ Given DE || BC ]
(i)
Now,
∠ADE + ∠CDE + ∠B = 180° ....[ ADB is a st. line ]
⇒ 64° + ∠CDE + 64° = 180°
⇒ ∠CDE = 180° − 128°
⇒ ∠CDE = 52°
(ii)
Given DE || BC and DC is the transversal.
⇒ ∠CDE = ∠DCB = 52° .......(ii)
Also, ∠ECB = 64° .....[ From (i) ]
But,
∠ECB = ∠DCE + ∠DCB
⇒ 64° = ∠DCE + 52°
⇒ ∠DCE + 64° − 52°
⇒ ∠DCE = 12°
Calculate x :
Let the triangle be ABC and the altitude be AD.
In ΔABD,
∠DBA = ∠DAB = 37° .....[Given BD = AD and angles opposite to equal sides are equal]
Now,
∠CDA = ∠DBA + ∠DAB .......[Exterior angle is equal to the sum of opp. interior angles]
∴ ∠CDA = 37° + 37°
⇒ ∠CDA = 74°
Now in ΔADC,
∠CDA = ∠CAD = 74° ....[ Given CD = AC and angles opposite to equal sides are equal]
Now,
∠CAD + ∠CDA + ∠ACD = 180°
⇒ 74° + 74° + x = 180°
⇒ x = 180° − 148°
⇒ x = 32°
Calculate x :
Let triangle be ABC and altitude be AD.
In ΔABD,
∠DBA = ∠DAB = 50° ...[Given BD = AD and angles opposite to equal sides are equal]
Now,
∠CDA = ∠DBA + ∠DAB ...[Exterior angle is equal to the sum of opp. interior angles]
∴ ∠CDA = 50° + 50°
⇒ ∠CDA = 100°
In ΔADC,
∠DAC = ∠DCA = x ...[Given AD = DC and angles opposite to equal sides are equal]
∴ ∠DAC + ∠DCA + ∠ADC = 180°
⇒ x + x + 100° = 180°
⇒ 2x = 80°
⇒ x = 40°
In the figure, given below, AB = AC.
Prove that: ∠BOC = ∠ACD.
Let ∠ABO = ∠OBC = x and ∠ACO = ∠OCB = y
In ΔABC,
∠BAC = 180° - 2x - 2y ....(i)
Since, ∠B = ∠C ...[AB = AC]
⇒ x = y
Now,
∠ACD = 2x + ∠BAC ...[ Exterior angle is equal to sum of opp. interior angles ]
∠ACD = 2x + 180° - 2x - 2y ...[ From(i) ]
∠ACD =180° - 2y ....(i)
In ΔOBC,
∠BOC = 180°- x - y
⇒ ∠BOC = 180°- y - y ...[ Already proved ]
⇒ ∠BOC = 180° - 2y ...(ii)
From (i) and (ii)
∠BOC = ∠ACD
In the figure given below, LM = LN; angle PLN = 110o.
calculate: (i) ∠LMN
(ii) ∠MLN
Given: ∠PLN = 110°
(i) We know that the sum of the measure of all the angles of a quadrilateral is 360°.
In quad. PQNL,
∠QPL + ∠PLN + ∠LNQ + ∠NQP = 360°
⇒ 90° + 110° + ∠LNQ + 90° = 360°
⇒ ∠LNQ = 360° − 290°
⇒ ∠LNQ = 70°
⇒ ∠LNM = 70° ........(i)
In ΔLMN,
LM = LN ........( Given )
∴ ∠LNM = ∠LMN ....... [angles opp. to equal sides are equal]
⇒ ∠LMN = 70° ....(ii) [ from(i) ]
(ii) In ΔLMN,
∠LMN + ∠LNM+ ∠MLN = 180°
But ∠LNM= ∠LMN = 70° .....[ From(i) and (ii)]
∴ 70° + 70° + MLN = 180°
⇒ ∠MLN = 180°− 140°
⇒ ∠MLN = 40°
An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠ CAB = 55o.
Find:
(i) ∠DCB
(ii) ∠CBD.
ln ΔABC,
AC = BC .......[Given]
∴ ∠CAB = ∠CBD ........[angles opp.to equal sides are equal]
⇒ ∠CBD = 55°
In ΔABC,
∠CBA + ∠CAB + ∠ACB = 180°
but, ∠CAB = ∠CBA = 55°
⇒ 55° + 55° + ∠ACB = 180°
⇒ ∠ACB = 180° − 110°
⇒ ∠ACB = 70°
Now,
In ΔACD and ΔBCD,
AC = BC .......[Given]
CD = CD ........[Common]
AD = BD .........[Given: CD bisects AB]
∴ ΔACD ≅ ΔBCD
⇒ ∠DCA = ∠DCB
⇒ ∠DCB =
⇒ ∠DCB = 35°.
Find x :
Let us name the figure as following :
In ΔABC,
AD = AC .......[ Given ]
∴ ∠ADC =∠ACD ......[Angles opp. to equal sides are equal]
⇒ ∠ADC = 42°
Now,
∠ADC = ∠DAB +∠DBA ...[Exterior angle is equal to the sum of opp. interior angles]
But,
∠DAB = ∠DBA ......[Given: BD = DA]
∴ ∠ADC = 2∠DBA
⇒ 2∠DBA = 42°
⇒ ∠DBA = 21°
For x:
x = ∠CBA + ∠BCA .......[Exterior angle is equal to the sum of opp. interior angles]
We know that,
∠CBA = 21°
∠BCA = 42°
∴ x = 21 + 42°
⇒ x = 63°
In the triangle ABC, BD bisects angle B and is perpendicular to AC. If the lengths of the sides of the triangle are expressed in terms of x and y as shown, find the values of x and y.
In ΔABD aand ΔDBC,
BD = BD ...[ Common ]
∠BDA = ∠BDC ...[ each equal to 90° ]
∠ABD = ∠DBC ...[ BD bisects ∠ABC ]
∴ ΔABD ≅ ΔDBC ...[ ASA criterion ]
Therefore,
AD = DC
x + 1 = y + 2
⇒ x = y + 1 ..... (i)
and AB = BC
3x + 1 = 5y - 2
Subtituting the value of x from (i)
3( y + 1 ) + 1 = 5y - 2
⇒ 3y + 3 + 1 = 5y - 2
⇒ 3y + 4 = 5y - 2
⇒ 2y = 6
⇒ y = 3
Putting y = 3 in (i)
x = 3 + 1
∴ x = 4
In the given figure; AE || BD, AC || ED and AB = AC. Find ∠a, ∠b and ∠c.
Let P and Q be the points as shown below:
Given:
∠PDQ = 58°
∠PDQ = ∠EDC = 58° .....[ Vertically opp . angles ]
∠EDC = ∠ACB =58° ........[ Corresponding angles ∵ AC || ED ]
In ΔABC,
AB = AC .......[ Given ]
∴ ∠ACB = ∠ABC = 58° ......[angels opp. to equal sides are equal]
Now,
∠ACB + ∠ABC+ ∠BAC = 180°
⇒ 58° + 58° + a = 180°
⇒ a = 180° − 116°
⇒ a = 64°
Since AE || BD and AC is the transversal.
∠ABC = b .......[ Corresponding angles ]
∴ b = 58°
Also since AE || BD and ED is the transversal
∠EDC = c .......[ Corresponding angles ]
∴ c = 58°
In the following figure; AC = CD, AD = BD and ∠C = 58o.
Find the angle CAB.
In ΔACD,
AC = CD ...[ Given]
∴ ∠CAD = ∠CDA
∠ACD = 58° ...[ Given ]
∠ACD + ∠CDA + ∠CAD = 180°
⇒ 58° + 2∠CAD = 180°
⇒ 2∠CAD = 122°
⇒ ∠CAD = ∠CDA = 61° ...(i)
Now,
∠CDA = ∠DAB + ∠DBA ...[ Ext. angel is equal to sum of opp. int. angles ]
But,
∠DAB = ∠DBA ...[ Given : AD = DB ]
∴ ∠DAB +∠ DAB = ∠CDA
⇒ 2∠DAB = 61°
⇒ ∠DAB = 30.5° ....(ii)
In ΔABC,
∠CAB = ∠CAD +∠DAB
∴ ∠CAB = 61° + 30.5°
⇒ ∠AB = 91.5°
In the figure given below, if AC = AD = CD = BD; find angle ABC.
In ΔACD,
AC = AD = CD ......[Given]
Hence, ACD is an equilateral triangle.
∴ ∠ACD = ∠CDA = ∠CAD = 60°
∠CDA = ∠DAB + ∠ABD ........[Ext angle is equal to the sum of opp. int. angles]
But,
∠DAB = ∠ABD ......[Given: AD = DB]
∴ ∠ABD + ∠ABD = ∠CDA
⇒ 2∠ABD = 60°
⇒ ∠ABD = ∠ABC = 30°
In triangle ABC; AB = AC and ∠A : ∠B = 8 : 5; find angle A.
Sol:
Let ∠A = 8x and ∠B = 5x
Given: AB = AC
⇒ ∠B = ∠C = 5x ...[Angles opp. to equal sides are equal]
Now,
∠A + ∠B + ∠C = 180°
⇒ 8x + 5x + 5x = 180°
⇒ 18x = 180°
⇒ x = 10°
Given that :
∠A = 8x
⇒ ∠A = 8 x 10°
⇒ ∠A = 80°.
In triangle ABC; ∠A = 60o, ∠C = 40o, and the bisector of angle ABC meets side AC at point P. Show that BP = CP.
SOl:
In ΔABC,
∠A = 60°
∠C = 40°
∴ ∠B = 180° - 60° - 40°
⇒ ∠B = 80°
Now,
BP is the bisector of ∠ABC.
∴ ∠PBC =
⇒ ∠PBC = 40°
In ΔPBC,
∠PBC = ∠PCB = 40°
∴ BP = CP ....[ Sides opp. to equal angles are equal.]
In triangle ABC; angle ABC = 90o and P is a point on AC such that ∠PBC = ∠PCB.
Show that: PA = PB.
Let PBC = PCB = x
In the right angled triangle ABC,
∠ABC = 90°
∠ACB = x
⇒ ∠BAC = 180° - ( 90° + x )
⇒ ∠BAC = ( 90°- x ) ...(i)
and
∠ABP = ∠ABC - ∠PBC
⇒ ∠ABP = 90° - x ...(ii)
Therefore in the triangle ABP;
∠BAP = ∠ABP
Hence, PA = PB ...[sides opp. to equal angles are equal]
ABC is an equilateral triangle. Its side BC is produced up to point E such that C is mid-point of BE. Calculate the measure of angles ACE and AEC.
SOl:
ΔABC is an equilateral triangle.
⇒ Side AB = Side AC
⇒ ∠ABC = ∠ACB ........[If two sides of a triangle are equal, then angles opposite to them are equal]
Similarly, Side AC = Side BC
⇒ ∠CAB = ∠ABC .......[If two sides of a triangle are equal, then angles opposite to them are equal]
Hence, ∠ABC = ∠CAB = ∠ACB = y(say)
As the sum of all the angles of the triangle is 180°.
∠ABC + ∠CAB + ∠ACB = 180°
⇒ 3y = 180°
⇒ y = 60°
∠ACB = ∠ACB = ∠ABC = 60°
Sum of two non-adjacent interior angles of a triangle is equal to the exterior angle.
⇒ ∠CAB + ∠CBA = ∠ACE
⇒ 60° + 60° = ∠ACE
⇒ ∠ACE = 120°
Now ΔACE is an isosceles triangle with AC = CF
⇒ ∠EAC = ∠AEC
Sum of all the angles of a triangle is 180°
∠EAC + ∠AEC + ∠ACE = 180°
⇒ 2∠AEC + 120° = 180°
⇒ 2∠AEC = 180° − 120°
⇒ ∠AEC = 30°
In triangle ABC, D is a point in AB such that AC = CD = DB. If ∠B = 28°, find the angle ACD.
Sol:
ΔDBC is an isosceles triangle.
As, Side CD = Side DB
⇒ ∠DBC = ∠DCB .......[If two sides of a triangle are equal, then angles opposite to them are equal]
And ∠B = ∠DBC = ∠DCB = 28°
As the sum of all the angles of the triangle is 180°
∠DCB + ∠DBC + ∠BCD = 180°
⇒ 28° + 28° + ∠BCD = 180°
⇒ ∠BCD = 180° − 56°
⇒ ∠BCD = 124°
Sum of two non-adjacent interior angles of a triangle is equal to the exterior angle.
⇒ ∠DBC+ ∠DCB = ∠DAC
⇒ 28° + 28° = 56°
⇒ DAC = 56°
Now ΔACD is an isosceles triangle with AC = DC
⇒ ∠ADC = ∠DAC = 56°
Sum of all the angles of a triangle is 180°
⇒ ∠ADC + ∠DAC + ∠DCA = 180°
⇒ 56° +56° + ∠DCA = 180°
⇒ ∠DAC = 180° − 112°
⇒ ∠DCA = 64° = ∠ACD.
In the given figure, AD = AB = AC, BD is parallel to CA and angle ACB = 65°. Find angle DAC.
We can see that the ΔABC is an isosceles triangle with Side AB = Side AC.
⇒ ∠ACB = ∠ABC
As ∠ACB = 65°
hence ∠ABC = 65°
Sum of all the angles of a triangle is 180°
∠ACB + ∠CAB + ∠ABC = 180°
65°+ 65° + ∠CAB = 180°
∠CAB = 180° − 130°
∠CAB = 50°
As BD is parallel to CA
Therefore, ∠CAB = ∠DBA since they are alternate angles.
∠CAB = ∠DBA = 50°
We see that ΔADB is an isosceles triangle with Side AD = Side AB.
⇒ ∠ADB = ∠DBA = 50°
Sum of all the angles of a triangle is 180°
∠ADB + ∠DAB + ∠DBA = 180°
50° + ∠DAB + 50° = 180°
∠DAB = 180° − 100° = 80°
∠DAB = 80°
The angle DAC is the sum of angle DAB and CAB.
∠DAC = ∠CAB + ∠DAB
∠DAC = 50°+ 80°
∠DAC = 130°
Prove that a triangle ABC is isosceles, if: altitude AD bisects angles BAC.
SOl:In ΔABC, let the altitude AD bisects ∠BAC.
Then we have to prove that the ΔABC is isosceles.
In triangles ADB and ADC,
∠BAD = ∠CAD ...(AD is bisector of ∠BAC)
AD = AD ...(common)
∠ADB = ∠ADC ....(Each equal to 90°)
⇒ ΔADB ≅ ΔADC ...(by ASA congruence criterion)
⇒ AB = AC ...(cpct)
Hence, ΔABC is an isosceles.
Prove that a triangle ABC is isosceles, if: bisector of angle BAC is perpendicular to base BC.
Sol:In Δ ABC, the bisector of ∠ BAC is perpendicular to the base BC. We have to prove that the ΔABC is isosceles.
In triangles ADB and ADC,
∠BAD = ∠CAD .......(AD is bisector of ∠BAC)
AD = AD ........(common)
∠ADB = ∠ADC .......(Each equal to 90°)
⇒ ΔADB ≅ ΔADC ......(by ASA congruence criterion)
⇒ AB = AC ........(cpct)
Hence, ΔABC is isosceles.
In the given figure; AB = BC and AD = EC.
Prove that: BD = BE.
In ΔABC,
AB = BC .......(given)
⇒ ∠BCA = ∠BAC .......(Angles opposite to equal sides are equal)
⇒ ∠BCD = ∠BAE ….(i)
Given, AD = EC
⇒ AD + DE = EC + DE ...(Adding DE on both sides)
⇒ AE = CD .....….(ii)
Now, in triangles ABE and CBD,
AB = BC .....(given)
∠BAE = ∠BCD ....[From (i)]
AE = CD ......[From (ii)]
⇒ ΔABE ≅ ΔCBD
⇒ BE = BD ....(cpct)
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