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Given: Points D and E are on side BC of ∆ABC, such that BD = CE and AD = AE. 

To prove: ∆ABD ≅ ∆ACE

Proof: 

In ∆ADE, seg AD = seg AE [Given] 

∴ ∠AED = ∠ADE …(i) [Isosceles triangle theorem] 

Now, ∠ADE + ∠ADB = 180° …(ii) [Angles in a linear pair] 

∴ ∠AED + ∠AEC = 180° ….(iii) [Angles in a linear pair] 

∴ ∠ADE + ∠ADB = ∠AED + ∠AEC [From (ii) and (iii)] 

∴ ∠ADE + ∠ADB = ∠ADE + ∠AEC [From (i)] 

∴ ∠ADB = ∠AEC ….(iv) [Eliminating ∠ADE from both sides] 

In ∆ABD and ∆ACE, seg BD ≅ seg CE [Given] 

∠ADB = ∠AEC [From (iv)] 

seg AD ≅ seg AE [Given] 

∴ ∆ABD ≅ ∆ACE [SAS test]

Given: In △ABC, AD ⊥ BC. 

R.T.P: AB² + CD² = BD² + AC² 

Proof: △ABD is a right angled triangle 

AB² – BD² = AD² ……. (1) 

△ACD is a right angle triangle

AC² – CD² = AD^{2............ (2)} 

From (1) and (2) 

AB² – BD² = AC² – CD² 

AB² + CD² = BD² + AC²

i) Similar figures: 

a) Any two circles 

b) Any two squares 

c) Any two equilateral triangles 

ii) Non-similar figures: 

a) A square and a rhombus 

b) A square and a rectangle

In ΔABC and ΔABD,

AC = AD (Given)

∠CAB = ∠DAB (AB bisects ∠A)

AB = AB (Common)

∴ ΔABC ≅ ΔABD (By SAS congruence rule)

∴ BC = BD (By CPCT)

Therefore, BC and BD are of equal lengths.

Given : In ABC, DE // AB AD = 8x + 9, CD = x + 3, 

BE = 3x + 4, CE = x 

By Basic proportionality theorem, 

If DE // AB then we should have

\(\frac{CD}{DA}=\frac{CE}{EB}\)

\(\frac{x+3}{8x+9}\) = \(\frac{x}{3x+4}\)

⇒ (x + 3) (3x + 4) = x (8x + 9) 

⇒ x (3x + 4) + 3 (3x + 4) – 8x² + 9x 

⇒ 3x² + 4x + 9x + 12 = 8x² + 9x 

⇒ 8x² + 9x – 3x² – 4x – 9x -12 = 0 

⇒ 5x² – 4x – 12 = 0 

⇒ 5x² – 10x + 6x – 12 = 0

⇒ 5x (x – 2) + 6 (x – 2) = 0 

⇒ (5x + 6) (x – 2) = 0 

⇒ 5x + 6 = 0 or x – 2 = 0 

⇒ x =\(\frac{-6}{5}\) or x = 2; 

x cannot be negative. 

∴ The value x = 2 will make DE // AB.

Given that AC = AD 

∠BAC = ∠BAD (∵ AB bisects ∠A)

Now in ΔABC and ΔABD 

AC = AD (∵ given) 

∠BAC = ∠BAD (Y given) 

AB = AB (common side) 

∴ ΔABC ≅ ΔABD 

(∵ SAS congruence rule)

In an isosceles triangle, two angles are always equal.

If one angle of a triangle is equal to the sum of other two, then the measure of that angle is 90°.

In an isosceles triangle, angles opposite to equal sides are equal.

Given 

ΔAHK ~ ΔABC, AK = 10 cm, 

BC = 3.5 cm and HK = 7 cm. 

We know that two triangles are similar if their corresponding sides are proportional.

⇒ \(\frac{AC}{AK}\) = \(\frac{BC}{HK}\)

⇒ \(\frac{AC}{10}\) = \(\frac{3.5}{7}\)

∴ AC = 5 cm

Correct answer is 40 m

Correct option is: (B) Δ FGH ~ Δ KIH

The difference between the lengths of any two sides of a triangle is smaller than the length of third side.

True

(c)Two triangles are similar if their corresponding sides are proportional. 

According to the statement: 

∆ABC~ ∆DEF 

if AB/ DE = AC/DF = BC/EF

(b) right-angled 

We have: 

AB² + BC² = 16² + 12² = 256 + 144 = 400 

and, AC² = 20² = 400 

∴ AB² + BC² = AC² 

Hence, ∆ABC is a right-angled triangle.

(d) 90°

Given: 

AC = BC 

AB² = 2AC² = AC² + AC² = AC² + BC² 

Applying Pythagoras theorem, we conclude that ∆ABC is right angled at C. 

Or, ∠C = 90°

Correct answer is (D) isosceles and similar

(b) DE/PQ = EF/RP 

In ∆DEF and ∆PQR, we have: 

∠D = ∠Q  and ∠R = ∠E

Applying AA similarity theorem, we conclude that ∆DEF ~ ∆QRP. 

Hence, DE/QR = DF/QP = EF/PR

Correct answer is (B) DE/PQ = EF/RP

Correct answer is (D) 100°

Correct answer is x = 20°

Correct answer is (A) 9

Correct answer is (B) DE = 12 cm, ∠F = 100°

Correct answer is (A)Δ PQR ~ Δ CAB

Correct answer is x = 80°, y = 75°

Correct answer is x = 60°, y = 120°, z = 30°

Correct answer is x = 30°, y = 40°. z = 110°

Given,

AB divides ∠DAC in the ratio 1: 3

∠DAB: ∠BAC = 1: 3

∠DAC + ∠EAC = 180°

∠DAC + 108° = 180°

∠DAC = 180° – 108°

= 72°

∠DAB = \(\frac{1}{4}\) x 72° = 18°

∠BAC = \(\frac{3}{4}\) x 72° = 54°

In ΔADB

∠DAB + ∠ADB + ∠ABD = 180°

18° + 18° + ∠ABD = 180°

36° + ∠ABD = 180°

∠ABD = 180° – 36°

= 144°

∠ABD + ∠ABC = 180°(Linear pair)

144° + ∠ABC = 180°

∠ABC = 180° – 144°

= 36°

In ΔABC

∠BAC + ∠ABC + ∠ACB = 180°

54° + 36° + x = 180°

90° + x = 180°

x = 180° – 90°

= 90°

Thus, x = 90°

Correct answer is x = 50°

In ∆ABC and ∆ADC,

AC = AC [common]

∠BAC = ∠DAC [given]

AB = AD [given]

∴ ∆ABC ≅ ∆ADC [SAS criterion]

∴ BC = DC [By C.P.C.T.] 

and ∠BCA = ∠DCA [By C.P.C.T.]

(i) ∆ADC ≅ ∆ABC

(ii) BC = DC

(iii) ∠BCA = ∠DCA

(iv) Line segment AC bisects ∠BAD and ∠BCD.

Correct answer is ∠S = 90°

(i) PQR, PRQ: 

∠TPQ = ∠PQR + ∠PRQ [Exterior angle property]

(ii) QPR, QRP: 

∠UQR = ∠QPR + ∠QRP [Exterior angle property]

(iii) RPQ, ROP: 

∠PRS = ∠RPQ + ∠ROP [Exterior angle property]

Correct answer is x = 70°, y = 80°

If QP || ST and QT is a transversal, then ∠PQR = ∠STU                      [alternate interior angles]

and if DS || RP and QT is a transversal, then ∠PRQ = ∠SUT              [alternate interior angles]

Hence, ∠S must be equal to ∠Pi.e. 90°.

PQR

∠ PRS = ∠ QPR + ∠ PQR

Given angles are 40°, 50°, 60° 

Sum of the angles = 40° + 50° + 60° 

= 150° <180° 

So, 40°, 50°, 60° cannot form a triangle.

Correct option is B) 50°

(d) 70^o, 60^o

We know that, the exterior angle is sum of interior opposite angles of triangle.

So, x + 50^o = 120^o

X = 120^o – 50^o

X = 70^o

We also know that, sum of angles of triangle are equal to 180^o.

So, 50^o + x + y = 180^o

50^o + 70^o + y = 180^o

120^o + y = 180^o

Y = 180^o – 120^o

Y = 60^o

Therefore, the measures of ∠x and ∠y is 70^o and 60^o respectively.

(d) 6 cm

We know that,

The sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

So, 6 cm + 10 cm > 3^rd side

3^rd side < 16 cm

The difference of the lengths of any two sides of a triangle is always smaller than the length of the third side.

So, 10 – 6 < 3^rd side

3^rd side > 4 cm

Therefore, 6 cm is the length of the 3^rd side.

(b) 4 cm

From Pythagoras theorem.

AC² = AB² + BC²

5² = AB² + 3²

AB² = 25 – 9

AB² = 16

AB = √16

AB = 4 cm

Given, 

y = 5x

According to the angle sum property of a triangle,

60° + x + y = 180°

60° + x + 5x = 180°

60° + 6x = 180°

6x = 180° - 60° = 120°

x = 120°/6 = 20°

y = 5x = 5 x 20 = 100°

According to the exterior angle property,

z = 60° + y

= 60° + 100°

= 160°

Third side must be 20 cm, because sum of two sides should be greater than the third side.

∴ Perimeter of the triangle

= Sum of all sides

= (9 + 20 + 20) cm

= 49 cm

We know that, corresponding parts of congruent triangles are equal.

(a) ΔSTU ≅ ΔDEF

∠S = ∠D, ∠T = ∠E and ∠U = ∠F ST = DE, TU = EF and SU = DF

(b) ΔABC ≅ ΔLMN

∠A = ∠L, ∠B = ∠M and ∠C = ∠N AB = LM, BC = MN and AC = LN

(c) ΔYZX ≅ APQR

∠T = ∠P, ∠Z = ∠Q and ∠X = ∠R YZ = PQ, ZX = QR and YX = PR

(d) ΔXYZ ≅ ΔMLN

∠X = ∠M, ∠Y = ∠L and ∠Z = ∠N XY = ML,YZ = LN and XZ = MN

(a) ΔABC ≅ ΔNLM

(b) ΔLMN ≅ ΔGHI

(c) ΔLMN ≅ ΔLON

(d) ΔZYX ≅ ΔWXY

(e) ΔOAB ≅ ΔDOE

(f) ΔSTU ≅ ΔSVU

(g) ΔPSR ≅ ARQP

(h) ΔSTU ≅ ΔPQR

i) From the figures, 

BE = CA 

EN = AR

BN = CR 

∴ S.S.S congruency is true. 

ii) From the figure, 

AL = SD = 4 cm

LD = LD = common side 

AD ≠ LS (3 cm ≠ 2.5 cm) 

∴ S.S.S congruency is not true.

Correct answer is (ii) ΔPQR ≅ ΔQPS is correct.

i) ∠P = ∠R 

∠T = ∠S

∠TQP = ∠SQR 

ii) ∠P=∠S 

∠Q = ∠R 

∠POQ = ∠SOR