Chapter-7, Triangles

Explanation:

The correct option is (c) SSA. SSA is not a criterion for congruence. If two triangles seem to be congruent by the SSA rule, then they cannot be said congruent.

Explanation:

The correct option is (b)∆CBA≅ ∆PRQ

In triangle ABC and triangle PQR

AB=QR,BC=PQ and CA=PR

Therefore,∆CBA ≅ ∆PRQ ( because SSS rule, “If all the three sides of one triangle are equivalent to the corresponding three sides of the second triangle, then the two triangles are said to be congruent”)

Explanation:

The correct option is  (b) 50°

Because in triangle ABC, AB=AC and∠B=50°

Here, AB=AC, and triangle ABC is an isosceles triangle.

Therefore,∠B=∠C

It is given that,∠B =50°

So, ∠C=50°

Explanation:

The correct option is 50°

Now, BC=AB

This implies, ∠A=∠C( Because the angles opposite to equal sides are equal)

Let the angles be x each, 

Now by using the angles sum property,

∠A+∠B+∠C=180°

 ⟹x+80°+x=180°

⟹2 180°-80°=100°

⟹2x=100°

⟹x=50°

Therefore, ∠A=50°

Explanation:

The correct option is (a) 4cm

It is given that in  ∆PQR,∠ R=∠P and QR=4cm and PR=5cm,

∆PQR,∠ R=∠P 

⟹ PQ=QR( the sides of the opposite  angles are equal)

⟹PQ=4 cm (because QR=4cm)

Therefore the length of PQ is 4 cm

Explanation:

The correct option is (b)BA> BD

It is given that, in triangle ABC, AD bisects  ∠BAC,

 because ∠BAD=∠CAD

In triangle ACD,∠BDA is an exterior angle

Therefore,∠BDA > ∠CAD( because exterior angle > interior opposite angles)

This implies,∠BDA > ∠BAD

Therefore, BA> BD (the side opposite to the greater angle is greater)

Explanation:

The correct option is (b) DF=5cm, ∠E=60° 

In triangle ABC, ∠A=80°and ∠ B=40°

So,  ∠A+∠B+∠C=180°

80+40+∠C=180°

So,∠C=60°

Now, triangle ABC = triangle FDE

The corresponding parts of the congruent triangles are congruent,

∠A=∠F=80°

∠B=∠D=40°

∠C=∠E=60°

And, AB=FD=5 cm

BC=DE 

AC=EF.

Explanation:

The correct option is (d) 3.4 cm

In a triangle, the difference between two sides should be less than the third side.

Here, the length of the third side will be 3.4 cm.

So, 3.4+1.5=4.9 lesser than 5 cm

We also know that in a triangle the sum of the length of two sides should be greater than the third side.

So, the length of the third side of a triangle cannot be 3.4 cm.

Explanation:

The correct option is (b)PQ> PR

In triangle PQR, if∠R is  > ∠ Q 

Then QR< PR.

It is given ∠R is  > ∠ Q 

This implies, PQ> PR……(side opposite to greater angle is longer)

Explanation:

The correct option is (a)Isosceles but not congruent 

In triangle ABC and triangle PQR,

 it is given that  ∠C=∠P and∠ B=∠Q,

So,   ∠A=  ∠R(Third angle of the triangle)

Thus,△  ABC∼ △  PQR

Also, it is given that AB=AC

So,∠ B=∠ C(isosceles triangle property)

But ∠ B=∠ Q &∠ C=∠ P

So, ∠ Q=∠ P or PR=QP.

Explanation:

The correct option is (b) AC=DE

We now that, foe a triangle to be congruent by the SAS axiom two of the side and the included angle of the triangles must be equal,

It is given that AB=DF &∠A=∠D

Now, here for triangle ABC=  triangle DEF by SAS axiom, we need AC=DE 

Hence proved option (b) AC=DE.

Explanation:

 It is given that, in △ABC &△DEF, AB=DE &AC=EF

We know that the two angles from the two triangles are equal so that the two triangles are congruent i.e,  ∠A=∠E

From SAS (Side Angle Side) criterion, we know that” if two sides of one triangle are respectively proportional to two corresponding sides of another, and if the included angles are equal, then the two triangles are congruent”

Therefore, △ABC =△DEF

Therefore, the two angles from the two triangles that must be equal so that the two triangles are congruent are ∠A and∠E.

Explanation:

 In triangle ABC and triangle DEF, it is given that,

∠ A=∠D

∠B=∠E

AB=EF

In triangle ABC, the two angles include the side given but in triangle DEF, the corresponding angles are equal but the side is not included in the angle. Hence, the triangles cannot be congruent.

Explanation:

It is given that,∠A=∠Q &∠B=∠R

AB lies between ∠A &∠B and side QR lies between ∠Q and ∠ R

So, AB should be equal to QR because they are corresponding sides.

According to the ASA rule, the two triangles are congruent.

Explanation:

It is given in the above that,∠A=∠Q  and∠B=∠R

The side BC lies opposite ∠Q 

So, the side RP lies opposite ∠Q 

Therefore BC should be equal to RP as they are the corresponding sides.

According to the AAS rule, the two triangles will be congruent.

Explanation:

 No, the above statement, If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent, is false.

Because by the congruent rule i.e, SAS rule “ the two sides and the included angle of one triangle are equal to the two sides and the included angles of the other triangle”

Explanation:

No, because the sides must be the corresponding sides. Two sides and an angle of a triangle are equal to two sides and an angle of another triangle. The two triangles are not equal because the corresponding equal angles must be contained on equal sides. 

The SAS criterion states that “if two sides of one triangle are respectively proportional to two corresponding sides of another, and if the included angles are equal, then the two triangles are congruent” Therefore both SAS and ASA are treated the same.

Explanation:

In the above, it is given that the sides are 4cm, 3cm, and 7cm

We know that the sum of two sides of a triangle is greater than the third side.

But it is given as, 4+3=7

So it is not possible to construct a triangle with the given lengths i.e 4cm, 3 cm and 7cm

Explanation:

No, we know that the two triangles are congruent if the sides and angles of one triangle are equal to the corresponding side and angles of the other triangle.

 It is given in the above that, triangle ABC≅ triangle RPQ

Here, we know that AB=RP, BC=PQ and AC=RQ

So,  BC cannot be equal to QR

So, the above statement is false.

Explanation:

Yes, if triangle PQR≅ triangle EDF, then it means that the corresponding angles and their sides are equal because we know that, two triangles are congruent, if the sides and angles of one triangle are equal to the corresponding sides and angles of another triangle

So, here, triangle PQR≅ triangle EDF

Therefore PO=ED and QR=DF  and PR=EF

So, it is true to say that PR=EF

Explanation:

It is given in the above that, triangle PQR, ∠P=70° &∠R=30°

The sum of the angles of a triangle is 180°

⟹∠P+∠Q+∠R=180°

⟹70°+30°+∠Q=180°

So, ∠Q=80°

Here ∠Q is the largest so, PR will be the longest side, this is because when the two sides of a triangle are unequal, then the angle opposite to the longer side is larger.

Explanation:

It is given in the above that, in triangle ABC, AD is the median

Here we know that the sum of any two sides of a triangle is greater than the third side

AB+BD>AD……….( a)

AC+DC>AD……….(b)

So it can be written as, AB+(BC+DC)+AC>2AD

So, AB+BC+AC>2AD is true.

Explanation:

Here, M is on line BC, so the perimeter of triangle ABC would be greater than 2 AM.

The sum of the two sides of a triangle will always be greater than the third side.

So, in triangle ABM, AB+BM>AM…..(a)

In triangle ACM, AC+CM>AM……(b)

Now let's add (a) and (b),

AB+BM+AC+CM>AM+AM

Therefore, AB+AC+BC>2AM

So, the perimeter of the triangle ABC is greater than 2 AM.

Explanation:

We know that in a triangle the sum of the two sides is greater than the third side. Above it is given 9cm,7 cm, and 17 cm as the lengths of the side.

So, 9+7=16cm < 17cm

Therefore, the sum of the two sides is less than the third side.

So, it is not possible to construct a triangle with lengths of its side as 9cm,7cm, and 17cm.

Explanation:

We know that, in a triangle, the sum of two sides is greater than the third side.

Above it is given that 8cm,7 cm, and 4cm as the lengths of the sides

So, 8+7=15 cm > 4cm

This means the sum of the two sides is greater than the third side.

So, it is possible to construct a triangle with the given lengths of the sides as 8cm,7cm, and 4cm.

Explanation:

In triangle PQS & triangle PRT,

It is given that PQ=PR and ∠Q=∠R

So,∠QPS=∠RPT(common angle)

Therefore, triangle  PQS ≅ triangle PRT(SAS rule).

Explanation:

It is given that, BC∥DA

Therefore, ∠CBO=∠DAO(Alternate interior angle)

And, ∠BCO=∠ADO(Alternate interior angle)

Also, it is given that, BC=DA

So, according to the ASA rule, triangle BOC≅ triangle AOD

Therefore, OB=OA and OC=OD, so that, O is the midpoint of both AB and CD.

Explanation:

In triangles PQR, It is given that PQ>PR

So, 1/2∠PRQ > 1/2∠PQR(angle opposite to longer side of a triangle is greater)

This implies,∠SRQ> ∠SQR(because RS is the bisector of∠PRQ &QS is the bisector of ∠PQR)

So, now in triangle SQR, ∠SRQ>∠SQR

Therefore, SQ>SR(because the side opposite to greater angles is longer).

Explanation:

Showing BD=CE

It is given in the above that ABC is an isosceles triangle with AB=AC, and BD and CE are its two medians. Now we have to prove that BD=CE.

In △ABD and△ACE, it is given that  AB=AC

2AE=2AD(D and E are midpoints)

So, AE=AD

Therefore,∠A=∠A(common)

So,△ABD≅ △ACE(by SAS rule)

Therefore BD=CE(by CPCT)

Hence proved that BD=CE.

Explanation:

It is given that D and E are the points on the side BC of such that BD=CE and AD=AE to show △ ABD≅ △ACE

We have proof that AD=AE

This implies,∠ADE=∠AED…..(a)[ since,angles opposite to equal sides are equal]

Now we have ∠ADB=∠ADE=180°[linear pair axiom]

This implies,∠ADB=180°-∠ADE

=180°-∠AED…..from(a)

In △ABD and△ ACE, ∠AD∠ACE[because∠AEC∠AED=180°, linear pair axiom]

BD=CE(given) and AD=AE(given)

Therefore,△ ABD≅ △ACE [by SAS  congruence rule].

Explanation: 

∠ADC=∠BCD=90°   —> Square

∠CDE=∠DCE=60°  —> equilateral_triangle

∴ ∠ADE=∠BCE=150° 

In △ADE and △BCE

AD=BC

DE=CE

∠ADE=∠BCE

△ADE = △BCE.

Explanation:

In △ABC and △DEF,

BA=ED and BF=EC

This implies, BF+FC=FC+EC

Therefore, BC=EF

∠BAC=∠EDF=90°

By using SAS congruence,△ ABC ≅△ DEF.

Explanation:

It is given in the above that, Q is a point on the side SR of a triangle PSR such that PQ=PR

 ⟹ PS>PR…..(a)

Proof: In△PQR, PQ=PR

 ⟹∠PQR=∠PRQ

So, by using equation (a)

∠PRQ>∠PSQ or ∠PRS> ∠PSR

In △PSR, ∠PRS> ∠PSR

⟹PS >PR

⟹PS >PQ hence proved.

Explanation:

We know that, in a triangle, the sum of any two sides is always greater than the third side.

Here in triangle PQS, we have,

PQ+QS>PS……..(a)

Also in triangle PSR, we have

PR+SR>PS……(b)

By adding (a) and (b)

PQ+QS+PR+SR>2PS

And also, PQ+PR+QR>2PS

So, PR=QS+SR.

Explanation:

It is given in the above that, ABC is a triangle and D is any point on the triangle AB=AC

Now we have to show that CD<BD

We know that “the angles opposite to the equal sides are equal”

So,∠ABC=∠ACB………..(a)

By considering the triangles ABC & DBC,

∠B is a common angle

Since ∠DBC is an internal angle of∠B

Therefore,∠ABC> ∠DBC

From(a), ∠ACB>  ∠DBC

We also know that in a triangle a side opposite to a greater angle is longer

BD>CD

Hence proved CD <BD.

Explanation:

In  triangle AMC and triangle BMD, we have

∠1=∠3( alternate∠s because l parallel to m)

∠2=∠4(vertical opposite ∠s)

AM=BM(GIVEN)

Therefore, triangle AMC=triangle BMD(by ASS congruence rule)

Therefore, CM=DM(CPCT)

So, M is also the midpoint of the CD

Explanation:

In triangle ABC, it is given that AB=AC

So we get that ∠ABC=∠ACB

Now by dividing both sides by 2, we get

1/2∠ABC=1/2∠ACB

So we get,∠OBC=∠OCB

Here let's use the exterior angle property,

∠MOC=∠OBC +∠OCB

We know that ∠OBC=∠OCB

So, ∠MOC=2∠OBC

We also know that OB is the bisector of∠ABC

Therefore, ∠MOC=∠ABC

Hence proved ∠MOC=∠ABC.

Explanation:

BD is the bisector of ∠ABC and CO is the bisector of ∠ACB which produces CB to D

 In ∆ABC, AB=AC

∠ACB=∠ABC

1/2∠ACB=1/2∠ABC

∠OCB=∠OBC

In triangle BOC,

∠OBC+OCB+ ∠BDC=180°

2∠OBC+∠BOC=180°

∠ABC+∠BOC=180°

∠ABC+∠OBA=180°

So, ∠DBA=∠BOC.

Explanation: 

∆ABC AD is  the bisector of ∠BAC 

∠1=∠2………………..(a)

Also in ∆ADC

∠3=∠2+∠C(exterior angle is equal to the sum of opposite interior angles)

∠3 > ∠2 (Exterior angle is greater than one of the interior angles)

But ∠1=∠2

And ∠3>∠1

Therefore AB>BD Hence proved.

Explanation:

In triangle ABD and Triangle ABC, we have BD=BC

 AB=AB(common)

 ∠ABD=∠ABC=90°

Therefore by using the SAS criterion of congruence we get ∆ABD≅ ∆ABC

 This implies AD=AC and ∠DAB=∠CAB(By CPCT)

 This implies AD = AC and ∠DAB = x[Therefore ∠CAB=x]

 Now ∠DAC=∠DAB+∠CAB=x+x=2x

Therefore ∠DAC=∠ACD

This implies DC=AD[side opposite to equal angles]

This implies 2BC=AD since DC=2BC

This implies 2BC=AC since AD=-AC Hence proved.

Explanation:

In triangles ABC and DEF ∠B=∠E and ∠C=∠F 

BC=EF

Now we have to prove that the triangle ABC and DEF are congruent 

Case 1: 

Let AB = DE

In Triangle ABC and DEF 

AB=DE

∠B=∠E

Given BC = EF

By the SAS criterion the triangles ABC and DEF are congruent 

Case 2 : 

             AB>DE

 Take a point P on AB such that PB = DE 

In triangles PBC and DEF 

PB = DE 

Given that ∠B = ∠E

Also given that BE=EF

By the SAS criterion, the triangles PBC and DEF are congruent 

By CPCTC, ∠PCB = ∠DEF 

It is given that ∠ ACB = ∠DEF

So∠PCB = ∠ACB 

This condition is satisfied only when P coincides with A

AB=DE

By Case 1 the triangles ABC and DEF are congruent 

Case 3 : 

 If AB < DE 

Take a point M on DE so that AB=ME

In triangle ABC and MEF, AB=ME is given ∠B=∠E

And also B 

BY CPCTC, ∠ACB = ∠MFE

It is given that ∠ ACB =∠DEF

So ∠MFE = ∠DEF this condition is satisfied only when M Coincides with D

So AB=DE

By case 1 the triangle ABC and DEF are congruent 

Therefore it is proved that the two triangles are congruent.

Explanation:

In triangle ABC, BD bisects ∠BAC and bisects the side BC

In ∆ABD  and ∆ACD,

 ∠BAD= ∠CAD(AD bisects ∠ BAC)

So, BD=CD(AD bisects side BC)

So, AD=AD (common)

Therefore ∆ABD≅ ∆ACD(SAS rule)

So, AB=AC(by CPCT)

Therefore ∆ABS is an isosceles triangle.

Explanation:

Let's project that QS to intersect PR at T

From triangle PQT, we have PQ+PT>QT

PQ+PT>SQ+ST……………(a)

ST+TR>SR…………(b)

Now let's add (a) and (b),

PQ+PT+ST+TR>SQ+ST+SR

PQ+PT+TR>SQ+SR

Therefore, PQ+PR>SQ+SR,

 Hence proved, SQ+SR< PQ+PR 

Explanation:

Given, triangle ABC is an equilateral triangle

We know that in an equilateral triangle, all the angles are equal.

∠A=∠B=∠C=x

The sum of the angles=180°

So,∠A+∠B+∠C=180°

x+x+x=180°

So,x=60°

Hence proved, ∠A=∠B=∠C=60°.

Explanation:

Let AB intersect LM at O. 

Now we have to prove that AO=BO

Here we know that ∠i=∠r……..(a)

[because the angle of incidence= angle of reflection]

∠B=∠i (corresponding angles).....(b)

Also ∠A=∠r(alternate interior angle).......(c )

From (a), (b),& (c ) we get that,

∠B=∠A

So, ∠BCO=∠ACO

In triangleBOC triangle AOC we have,

∠1=∠2(each 90 degrees)

Therefore OC=OC (common side)

So, ∠BCO=∠ACO(As proved in the above)

So, triangle BOC≅ triangle AOC(ASA congruence rule)

Hence proved AO=BO(CPCT).

Explanation:

In triangle ABD & triangle ADC,  it is given that ∠ADB=∠ADC

According to the question, AB=BC &AD=AD(common side)

By RHS criterion of congruence,

We have, ∆ABD≅ ∆ACD

∠BAD=∠CAD(CPCT)Hence proved

Explanation:

Here we have to prove that BPQ is an isosceles triangle

We know that BP is the bisector of ∠ABC

So∠1=∠2……(a)

PQ is parallel to BA and BP cuts them

Therefore,∠1=∠3(alternate angles).....(b)

From equations (a) and(b)

We get that,∠2=∠3

In triangle BPQ, we have ∠2=∠3

so, PQ=BQ

Therefore it is proved that BPQ is an isosceles triangle.

Explanation:

In ∆ABC &∆CBD, it is given that  AB=BC & AD=CD.  

And BD=BD(common side)

So, ∆ABC ≅∆CBD( according to the SSS congruence rule)

This implies,∠1=∠2(by CPCT)

And ∠3=∠4

Therefore, BD bisects both ∠ABC and∠ADC.

Explanation:

It is given in the above that ABC is a right triangle and AB=AC

Since, the hypotenuse is the longest side, and here BC is a hypotenuse

Here,∠BAC=90 degree

In triangle CAD & triangle BAD, we have AC=AB

Since AD is the bisector of∠A, 

∠1=∠2

so,AD=AD(common side)

According to the SAS criterion of congruence,

We get that,∆CAD≅∆BAD so, CD=BD(CPCT)

 Since the Midpoint of the hypotenuse of a right triangle is equidistant from the 3 vertices of a triangle

AB=BD=CD….(a)

now, BC=BD+CD

So, BC=AD+AD (using equation a)

Therefore BC=2AD.

Explanation:

In square ABCD, ∠1 & ∠2=90 degrees

So ∠1=∠2………(a)

In triangle OAB, we have ∠3 and ∠4 which are equal to 60 degrees.

So ∠3=∠4………(b)

Now let's subtract (b) from (a)

∠1-∠3=∠2-∠4

So, ∠5=∠6

In triangle DAO& triangle CBO, it is given that AD=BC

So,∠5=∠6(proved above)

OA=OB

By SAS criterion of congruence,

We have, triangle DAO≅  triangle CBO

So OD=OC

This implies triangle OCD is an isosceles triangle.

Explanation:

In∆ABD and∆ACD, it is given that AB=AC

So, AD=AD(common side)

And also it is given that DB=DC

Therefore ∆ ABD ≅ ∆ACD(by SSScongruence rule)

This implies,∠BAD=∠CAD(by CPCT)

That is,∠BAO=∠CAO

In ∆AOB&∆AOC, it is given AB=AC

AO=OA(common side)

And ∠BAO=∠CAO(proved above)

Therefore,∆AOB  ≅ ∆AOC(by SAS congruence rule)

so,BO=OC(by CPCT)

And,  ∠AOB= ∠AOC(by CPCT)....(a)

But  ∠AOB+ ∠AOC=180 degrees(linear pair axiom)

so, ∠AOB ∠AOB=180 degrees(from equation a)

That is, 2 ∠AOB=180 degrees

 ∠AOB=180/2=90 degrees

Therefore AD⊥BC & AD bisects BC i.e, AD  is the perpendicular bisector of BC.

Explanation:

In ∆ADC &∆ BEC, 

We have, AC=BC(given)....(a)

Since, ∠ADC & ∠BEC=90 degrees

 ∠ADC= ∠BEC

 ∠ACD= ∠BCE(common angle)

So,  ∆ADC ≅ ∆ BEC,(SSS congruence)

CE=CD…..(b) [CPCT]

Now by subtracting b from a

We get, AC-CE=BC-CD

This implies, AE=BD, hence proved.

Explanation:

It is given that in triangle ABC, AD is the median drawn from A to BC

Now we have to prove that AB+AC is greater than AD

Let's produce AD to E so that DE=AD,

Join BE

Proof: in triangle ADC and triangle EDB we have AD=DE (constant)

DC=BD as D is the midpoint

 ∠ADC= ∠EDB(vertically opposite angles)

Therefore in ∆ABE,∆ADC≅∆EDB by SAS rule

So, this gives BE=AC

AB+BE >AE

AB+AC>2AD because  ,AD=DE &BE=AC

So, the sum of any two sides of a triangle is greater than twice the median with respect to the third side.

Explanation:

Here join the diagonals AC & BD of the quadrilateral

Now consider the triangle OAB,

We know that the sum of 2 sides of a triangle is greater than the third side

so, OA+OB> AB…….(a)

Now let's consider the triangle OBC

We know that the sum of the 2 sides of a triangle is greater than the third side.

OB+OC>BC…….(b)

Let's consider triangle OCD,

OC+OD>CD………..( c)

And in triangle ODA,

OD+OA>DA………(d)

Now let's add equations a,b,c and d

OA+OB+OB+OC+OC+OD+OD+OA>AB+BC+CD+DA

So, 2(OA+OB+OC+OD)>AB+BC+CD+DA

From the above figure, AC=OA+OC

And BD=OB+OD

So,2[(OA+OC)+(OB+OD)+]>AB+BC+CD+DA

2(AC+BD)>AB+BC+CD+DA

Therefore the above expression can also be rewritten as,

AB+BC+CD+DA <2(BD+AC)

Hence proved,

AB+BC+CD+DA<2(BD+AC).

Explanation:

It is given that ABCD is a quadrilateral

Now we have to show that AB+BC+CD+DA>AC+BD

Let's join the diagonals AC & BD of the quadrilateral

Considering triangle ABC,

We know that the sum of two sides of a triangle is greater than the third side

AB+BC>AC……….(a)

Considering triangle BCD

BC+CD>BD…….(b)

Considering triangle CDA

CD+DA>AC……( c)

Considering triangle DBA,

DA+AB>BD……..(d)

Now let's add a,b,c, and d equations

AB+BC+BC+CD+CD+DA+DA+AB>AC+BD+AC+BD

On rearranging we get that,

AB+AB+BC+BC+CD+CD+DA+DA>AC+AC+BD+BD

That is, 2(AB+BC+CD+DA)>2(AC+BD)

Therefore, AB+BC+CD+DA>AC+BD.

Explanation:

It is given that ABC is a triangle D is the midpoint of side AC

BD=1/2AC……(a)

Now we have to show that,∠ABC is a right angle

Here D is the midpoint of AC

so, AD=CD

AC=AD+CD

Now, AC=AD+AD or CD+CD

So, AC=2AD or 2CD

so,AD=CD=1/2AC……..(b)

Now by comparing equations a and b

AD=CD=BD……(c )

By considering triangle DAB

From(c )AD=BD,

We know that the angles opposite to the equal sides are equal

∠ABD=∠BAD…….(d)

Considering triangle DBC,

From (c )BD=CD

We know that the angles opposite to the equal sides are equal

∠BCD=CBD……..(e)

Considering triangle ABC,

By angle sum property,

∠ABC+ ∠BAC+ ∠ACB=180° 

From the figure, ∠BAC=∠BAD

∠AC∠DCB

now,∠ABC+∠BAD+∠BCD=180° 

From equations d and e,

∠ABC+ ∠ABD+ ∠CBD=180° 

Given, BD=1/2AC

∠ABC =∠ABD  +∠CBD

Now, ∠ABC+∠ABC=180° 

2∠ABC=180° 

∠ABC=180/2=90° 

Hence proved,∠ABC=90°.

Explanation:

Let P be the midpoint of the hypotenuse of the right triangle ABC right angled at B

Now let's draw a line parallel to BC from P meeting B at O

Join PB

In triangle PAD & triangle PBD,

∠PDA=∠PDB is 90° each due  to the convo of mid-point theorem

so,PD=PD(common)

AD=DB….(D is midpoint of AB)

So, triangle PDA & PDB are congruent by the SAS rule

PA=PB….(CPCT)

As, PA=PC( P is midpoint)

Therefore, PA=PC=PB.

Explanation:

It is given in the above that the lines l, m, and n intersect at point O, P is a point on line n and such that P is equidistant from l&n

In triangleOQP & triangle ORP, we have,

∠1=∠2(because each equal to 90 degrees)

Therefore, OP=OP(common side)

It is given that, PQ=QR

So, by the RHS criterion of congruence, we have

∆ OQP≅∆ ORP

therefore,∠3=∠4(CPCT)

So, is a bisector of ∠QOR, hence proved.

Explanation:

:

Let's construct AN and BN at point N

Let's consider triangles ANM & BNM,

Here we know that N is the midpoint of the line AB

So we get, AM=BM

From the above figure we know that,

∠ANM= ∠BNM =90 degrees

MN=MN(common)

By the SAS congruence criterion, triangle ANM≅ triangle BNM

so,AN=BN(CPCT)......(1)

∠ANM=∠BNM(CPCT)

Let's subtract LHS & RHS by 90 degrees

90° -∠ANM=90°-∠BNM

∠AND=∠BNC…..(2)

Now consider triangle AND & triangle BNC

AN=BN

∠AND=∠BNC

Here we know that N is the midpoint of the line dc

DN=CN

By SAS congruence, triangle AND≅ triangle BNC

AD=BC(CPCT)

Hence proved, AD=BC.

Explanation:

In triangle ABC and triangle ADC,

∠BAC=∠DAC(given)

∠ACB=∠ACD

Therefore AC=AC(common)

Therefore triangle AB ≅  triangle ADC(ASAaxiom)

So, AB=AD & CB=CD(CPCT)

Explanation:

Construct a triangle ABC where DE is perpendicular to BC

Consider the △ DAC and △ DEC

We know that

∠ BAC = ∠ DAC = 90°

From the figure, we know that CD bisects ∠ C

So we get

∠ DCA = ∠ DCE

We know that CD is common i.e. CD = CD

By AAS congruence criterion

△ DAC ≅ △ DEC

So we know that DA = DE …… (1)

AC = EC (c. p. c. t) ….. (2)

It is given that AB = AC

We know that the angles opposite to equal sides are equal

∠ B = ∠ C

We know that the sum of angles of △ ABC is 180°

∠ A + ∠ B + ∠ C = 180°

By substituting the values

90° + ∠ B + ∠ B = 180°

On further calculation

2 ∠ B = 180° - 90°

2 ∠ B = 90°

By division

∠ B = 45°

Considering the △ BED

We know that ∠ BED = 90°

So we can write it as

∠ BDE + ∠ B = 90°

By substituting the values

∠ BDE + 45° = 90°

On further calculation

∠ BDE = 90° - 45°

By subtraction

∠ BDE = 45°

It can be written as

∠ BDE = ∠ DBE = 45°

We know that DE and BE are the equal sides of an isosceles triangle

DE = BE …….. (3)

By comparing equations (1) and (3)

We get

DA = DE = BE …… (4)

We know that BC = BE + EC

By considering the equations (ii) and (iv)

We get

BC = DA + AC

We can also write it as

AC + AD = BC

Therefore, it is proved that AC + AD = BC.

Explanation:

Construction: join BD

Now in ∆ABD, AD is greater than AB(since ABis smallest side)

So,∠1 is greater than ∠3( angle opposite to larger side is greater)......(1)

In ∆ BCD, CD is greater than BC(CD is the largest side)

This implies, ∠2 is greater than ∠4(angle opposite to larger side is greater).....(2)

On adding equations 1 and 2

We get that,∠1+∠2 is greater than ∠3+∠4

Hence proved,∠B Is greater than ∠D.

Explanation:

Consider ∆ PQR where PR is the longest side

So we get PR> PQ i,e, ∠Q > ∠R…..(1)

We also know that PR> QR i,e,∠Q > ∠P…..(2)

By adding both equations,

∠Q+∠Q >  ∠R+∠P

By adding ∠Q on both LHS & RHS

2∠Q+∠Q  > ∠R+∠P+∠Q

We know that, ∠R+∠P+∠Q=180 degree

So we get, 3∠Q  >180°

By division, ∠Q  > 2/3 (90°)

I.e., ∠Q > 2/3 of a right angle.

Therefore, it is proved that in a triangle, other than an equilateral triangle, the angle opposite the longest side is greater than 2/3 of a right angle.

Explanation: 

In △AOB and △AOD

AB=AD (given)

∠BAD=∠DAO (AC bisects ∠A)

∴AO=AO [common]

⇒△AOB≅△AOD [SAS]

DO=BO.....(i) [CPCT]

∠AOD=∠AOB

∠AOD + ∠AOB=180°

⇒∠AOD=∠AOB=90°.....(ii)

From (i) and (ii) 

AC is perpendicular bisector of BD