∠B = 40°, ∠A = 70°     ... (Given)

In Δ ABC,

∠ACD = ∠A + ∠B          ….(Remote interior angle theorem)

= 70° + 40° = 110°

Hence, the measure of ∠ACD is 110°.

In Δ PQR,

∠P + ∠Q + ∠R = 180° ... (The sum of all angles of a triangle is 180°.)

70° + 65° + ∠R = 180°

∴ 135° + ∠R = 180°

∴ ∠R = 180° - 135° = 45°

Hence, the measure of ∠R is 45°.

The measures of the angles of a triangle are x°, (x - 20)°, (x - 40)°.

x° + (x - 20)° + (x - 40)° = 180°   ... .(The sum of the angles of a triangle is 180°.)

∴ 3x – 60 = 180

∴ 3x = 180 + 60

∴ 3x = 240

∴ x = 240/3

∴ x = 80

Now, x° = 80°

(x - 20)° = (80 - 20)° = 60°

(x - 40)° = (80 - 40)° = 40°

Hence, the measures of the angles of the triangle are 80°, 60° and 40°.

Let us suppose the angles of a Δ PQR such that ∠P < ∠Q < ∠R.

∴ ∠Q = 2∠P,

∴ ∠R = 3∠P

Now, ∠P + ∠Q + ∠R= 180°    ... (The sum of the angles of a triangle is 180°.)

∠P + 2∠P + 3∠P = 180°

∴ 6∠P = 180°

∠P = 180°/6 = 30°

∴ P = 30°

∠Q = 2∠P = 2 x 30° = 60°

∠R = 3∠P = 3 × 30° = 90°

Hence, the measure of each angle is 30°, 60° and 90° respectively. P = 30°

∠NEM + ∠NET = 180°           ... (Linear angle property)

∴ y + 100° = 180°

∴ y = 80°

Also, ∠NME + ∠EMR = 180°   ... (Linear angle property)

∴ z + 140° = 180°

∴ z = 40°

Now, In Δ NEM

∠N + ∠E + ∠M = 180° ... (Angle sum property)

∴ x + y + z = 180°

∴ x + 80° + 40° = 180°

∴ x + 120° = 180°

∴ x = 60°

Hence, the values of x, y and z are 60°, 80° and 40° respectively.

AB || DE and AD is a transversal line.

∠BAR = ∠RDE = 70°              .... (Alternate angles)

And line AD is their transversal

∴ ∠BAD ≅ ∠EDA                    ... (Alternate angles)

∠BAD = 70°                          .. (Given)

:. ∠EDA = 70°

In other words ∠EDR = 70°   ... (A-R-D)

∠ARE, ∠RDE exterior angle

∴ ∠ARE = ∠EDR + ∠DER        ... (remote interior angle theorem)

∴ ∠ARE = 70° + 40°

∴ ∠ARE = 110°

∠ARE + ∠DRE = 180°            ... (angles in linear pair)

∴ 110° + ∠DRE= 180°

∴ ∠DRE = 180° - 110°

∴ ∠DRE = 70°

∠OAB ≅ ∠OAC ... (Seg AO bisects 4BAC) ... (i)

∠OBA ≅ ∠OBC ... (Seg RO bisects ZABC) ... (ii)

In Δ ABC,

∠BAC + ∠ABC + ∠ACB = 180° ... (The sum of the angles of a triangle is 180°.)

∴ ∠BAC + ∠ABC + 70° =1 80°

∴ ∠BAC + ∠ABC = 180° - 70°

∴ ∠BAC + ∠ABC = 110°

\(\frac{1}{2}\)(∠BAC) + \(\frac{1}{2}\)(∠ABC) = \(\frac{1}{2}\) x 110° ... [Multiplying both sides by \(\frac{1}{2}\) ]

∴ ∠OAB + ∠OBA = 55° ... (iii) ... [From (i) and (ii)]

In Δ OAB,

∠OAB + ∠OBA + ∠AOB = 180° ... (The sum of the angles of a triangle is 180°.)

∴ 55° + ∠AOB = 180° ….[From (iii)]

∴ ∠AOB = 180° -55°

∴ ∠AOB = 125°

Given: line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively.

To prove: m∠PTQ = 90°

Proof:

∠TPB = ∠TPQ = \(\frac{1}{2}\)∠BPQ         ... (i) ….[Ray PT bisects ∠BPQ]

∠TQD = ∠TQP = \(\frac{1}{2}\)∠PQD        ... (ii) …[Ray QT bisects ∠PQD]

line AB || line CD and line PQ is their transversal.      ... (Given)

∴ ∠BPQ + ∠PQD = 180°                 ... [Interior angles]

\(\frac{1}{2}\)(∠BPQ) + \(\frac{1}{2}\)(∠PQD) = \(\frac{1}{2}\) x 180°    ... [Multiplying both sides by \(\frac{1}{2}\) ]

∴ ∠TPQ + ∠TQP = 90° ... (iii)

In Δ PTQ,

∠TPQ + ∠TQP + ∠PTQ = 180°         ... (The sum of the angles of a triangle is 180°.)

∴ 90° + ∠PTQ = 180°            ... [From (iii)]

∴ ∠PTQ = 180° - 90°

∴ ∠PTQ = 90°

∴ m∠PTQ = 90°

Hence proved.

∠b = 70°                              ... (Vertically opposite angles)

∠c + 100° =180°                  ... (Angles in a linear pair)

∴ ∠c = 180° - 100°

∴ ∠c = 80°

∠a + ∠b + ∠c = 180°            ... (The sum of the angles of a triangle is 180°.)

∴ ∠a + 70° + 80° = 180°

∴ ∠a + 150° = 180°

∴ ∠a = 180° - 150°

∴ ∠a = 30°

Hence ∠a = 30°, ∠b = 70°, ∠c = 80°.

Proof :

∠DEG ≅ ∠GEF                      ... (Ray EG bisects ZDEF)

Let ∠DEG = ∠GEF = x°          ... (1)

∠DEF = ∠DEG + ∠GEF           ... (Angles addition postulate)

∴ ∠DEF = x° + x°                  ... [From (1)]

∴ ∠DEF = 2x°                       ... (2)

Line DE || line GF and line EF is the transversal.

∠DEF ≅ ∠GFM                      ... (Corresponding angles)

∴ ∠GFM = 2x°                      ... [From (2)] ... (3)

∠DFG ≅ ∠GFM                      ... (Ray FG bisects 4DFM)

∴ ∠DFG = 2x°                       ... [From (3)] ... (4)

∠DFM = ∠DFG + ∠GFM          ... (Angles addition postulate)

∴ ∠DFM = 2x° + 2x°             ... [From (3) and (4)]

∴ 4DFM = 4x°                      ... (5)

∠DFM is an exterior angle of Δ DEF.

∴ ∠DFM = ∠EDF + ∠DEF                ... (Theorem of remote interior angles)

∴ 4x° = ∠EDF + 2x°

∴ ∠EDF = 4x° - 2x°

∴ ∠EDF = 2x°

∴ ∠EDF = 2∠DEG                  ... [From (1)]

∴ ∠DEG = \(\frac{1}{2}\)∠EDF

Line DE || line GF                 ... (Given)

and line EG is the transversal.

∠DEG ≅ ∠FGE                      ... (Alternate angles)

∴ ∠FGE = x°                         ... (6)

In Δ FEG,

∠FGE ≅ ∠GEF                       ... [From (6) and (1)]

∴ EF = FG                            ... (Isosceles triangle theorem)

By SSS test  Δ ABC ≅ Δ PQR

By SAS test Δ XYZ ≅ Δ LMN

By ASA test Δ PRQ ≅ Δ STU

By hypotenuse side test Δ LMN ≅ Δ PTR

From the information shown in the figure,

in Δ ABC and Δ PQR

∠ABC ≅ ∠PQR

seg BC ≅ seg QR

∠ACB ≅ ∠PRQ

∴ Δ ABC ≅ Δ PQR         .......[ASA] test

∴ ∠BAC ≅ [∠QPR ]       .......corresponding angles of congruent triangles.

seg AB ≅ [seg PQ] and [seg AC ] ≅ seg PR     ….. corresponding sides of congruent triangles

From the information shown in the figure,,

In Δ PTQ and Δ STR

seg PT ≅ seg ST

∠PTQ ≅ ∠STR               ....vertically opposite angles

seg TQ ≅ seg TR

∴ Δ PTQ ≅ Δ STR ....... [SAS] test

∴ ∠TPQ ≅ [∠TSR] and [∠TQP] ≅ ∠TRS    ….. corresponding sides of congruent triangles

seg PQ ≅ [seg SR]      ….corresponding sides of congruent triangles.

In Δ ABC and Δ QPR

AB = PQ,  BC= PR        ... (Given)

∠BAC = ∠PQR = 90° ... (Given)

∴ Δ BAC ≅ Δ PQR         ... [Hypotenuse side test]

∴ seg AC ≅ seg QR       ... [c.s.c.t.]

∠ABC ≅ ∠QPR and ∠ACB ≅ ∠QRP ... [c.a.c.t.]

Δ LMN ≅ Δ PNM           ... (SSS test of congruence)

∠MLN ≅ ∠NPM              ... [c.a.c.t.]

∠LMN ≅ ∠PNM              ... [c.a.c.t.]

∠LNM ≅ ∠PMN              ... [c.a.c.t.]

In Δ ABD and Δ CBD,

seg BA ≅ seg BC          ...(Given)

seg AD ≅ seg CD         ...(Given)

seg BD ≅ seg BD          ... (Common side)

∴ Δ ABD ≅ Δ CBD        ... (SSS test of congruence)

Proof : In Δ PQT and Δ RQS,

∠PQT ≅ ∠RQS              ... (Common angle)

seg PQ ≅ seg RQ          ...(Given)

∠QPT ≅ ∠QRS              ...(Given)

∴ Δ PQT ≅ Δ RQS        ... (ASA test of congruence)

In Δ ABC, AB = AC

∴ ∠ABC = ∠ACB           ... (Isosceles triangle theorem)

∴ ∠ACB = 50°              ... (given)

∴ ∠ABC = 50° i.e. x = 50°

In Δ BDC, DB = DC

∴ ∠DBC = ∠DCB           ... (Isosceles triangle theorem)

∠DBC = 60°                 ... (given)

∠DCB = 60° i.e. y = 60°

Now, ∠ABD = ∠ABC + ∠DBC ... (Angle addition property)

= 50° + 60°

∴ ∠ABD = 110°

Also, ∠ACD = ∠ACB + ∠DCB ... (Angle addition property)

= 50° + 60°

= 110°

∴ ∠ACD = 110°

Hence, the values of x = 50° and y = 60°, ∠ABD = 110° and ∠ACD = 110° respectively.

In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse.

∴ length of median of its hypotenuse = \(\frac{1}{2}) x 15 = 7.5

Hence, the length of median of its hypotenuse is 7.5 units.

In Δ PQR,

∠PQR = 90°        ... (Given)

By Pythagoras theorem,

PR² = PQ² + QR²

∴ PR² = 12² + 5²

∴ PR² = 144 + 25

∴ PR² = 169

∴ PR = \(\sqrt{169}\)

PR = 13

In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse.

∴ QS = \(\frac{1}{2}) x PR

∴ QS = \(\frac{1}{2}) x 13

∴ QS = 6.5

Hence, l(QS) = 6.5 units.

GT = 2.5             ... (Given)

PG : GT = 2 : 1   ….(The centroid divides each median in the ratio 2:1)

∴ \(\frac{PG}{GT}) = \(\frac{2}{1})

∴ \(\frac{PG}{2.5}) = \(\frac{2}{1})

∴ PG = 2 x 2.5

∴ PG = 5

Now, PT = PG + GT = 5 + 2.5

∴ PT = 7.5 units

Hence, the length of PG and PT is 5 and 7.5 units respectively.

Here, AX = 2 cm ... (Given)

Point A lies on the bisector of XYZ. ... [Given]

Point A is equidistant from the sides of ∠XYZ.     ... [Every point on the bisector of an angle is equidistant from the sides of the angle]

∴ AZ = AX

∴ AZ = 2 cm

Hence, the length of AZ is 2 cm.

seg PT ⊥ ray ST           ... (Given)

seg PR ⊥ ray SR           ... (Given)

seg PR ≅ seg PT           ... (Given)

i.e. PR = PT

∴ point P is equidistant from the sides of ∠TSR

∴ point P lies on the bisector of ∠TSR      ... (Angle bisector theorem) ... (1)

∠RST = 56°

∴ ∠RSP = \(\frac{1}{2}\) ∠RST                  ... [From (1)]

∴ ∠RSP = \(\frac{1}{2}\) x 56°

Hence, ∠RSP = 28°.

PQ = 10 cm, QR = 12 cm, PR = 8 cm      ... (Given)

∴ QR > PQ > PR

∴ ∠QPR > ∠PRQ > ∠PQR        ... (Angle opposite to greater side is greater)

Hence, ∠QPR is the greatest angle and ∠PQR is the smallest angle.

In Δ FΑΝ,

∠F + ∠A + ∠N = 180°           ... (sum of measures of angles of a triangle is 180°)

80° + 40° + ∠N = 180°

120° + ∠N = 180°

∴ ∠N = 180° - 120° = 60°

Now, in Δ FAN, ∠F and ∠A are the largest and smallest angles.

The side opposite to the largest angle is the greatest side and the side opposite to the smallest angle is the smallest side.

Hence, AN is the greatest side of the triangle and FN is the smallest side of the triangle.

Consider an equilateral triangle ABC.

In Δ ABC,

AB ≅ BC

∴ ∠ACB = ∠BAC           ... (1) ... (Angles opposite to equal sides)

In Δ ABC,

AB ≅ CA

∴ ∠ACB = ∠ABC           ... (2) ... (Angles opposite to equal sides)

From (1) and (2), we get

∠BAC ≅ ∠ABC ≅ ∠ACB

Hence, an equilateral triangle is equiangular.

Proof : In Δ ADB and Δ ADC.

∠BAD ≅ ∠CAD                       ... (Given)

seg AD ≅ seg AD                  ... (Common side)

∠ADB ≅ ∠ADC                      ... (Each measures 90°)

∴ Δ ABD ≅ Δ ADC                 ... (ASA test of congruence)

∴ seg AB ≅ seg AC                ... (c.s.c.t.)

∴ Δ ABC is isosceles.

In Δ PQR,

seg PR ≅ seg PQ          ... [Given]

∴ ∠PQR ≅ ∠PRQ           ... (i) [Isosceles triangle theorem]

∠PRQ is the exterior angle of APRS.

∴ ∠PRQ > ∠PSR            ... (ii) [Property of exterior angle]

∴ ∠PQR > ∠PSR            ... [From (i) and (ii)]

i.e. ∠Q > ∠S                ... (iii)

In Δ PQS,

∠Q > ∠S                      ... [From (iii)]

∴ PS > PQ                   ... [Side opposite to greater angle is greater]

∴ seg PS > seg PQ

Given: In Δ ABC, seg AD and seg BE are altitudes, and AE = BD.

To prove: seg AD ≅ seg BE.

Proof:

In Δ ADB and Δ BEA,

∠ADB = ∠BEA = 90°     ... [Given]

seg BD ≅ seg AE          ... [Given]

seg AB ≅ seg BA          ... [Common side]

By hypotenuse side test,

∴ Δ ADB ≅ Δ BEA

∴ seg AD = seg BE                ... [c.s.c.t.]

Hence proved.

Consider, Δ XYZ ≅ Δ LMN

∴ ∠X ≅ ∠L, ∠Y ≅ ∠M and ∠Z ≅ ∠N   ... (Corresponding angles of similar triangles)

\(\frac{XY}{LM}\) = \(\frac{YZ}{MN}\) = \(\frac{XZ}{LN}\)   ... (Corresponding sides of similar triangle)

Consider, Δ XYZ ~ Δ PQR

\(\frac{XY}{PQ}\) = \(\frac{YZ}{QR}\) = \(\frac{XZ}{PR}\) ... (Corresponding sides of similar triangle)

∴ \(\frac{4}{8}\) = \(\frac{6}{QR}\) = \(\frac{5}{PR}\)

∴ \(\frac{4}{8}\) = \(\frac{6}{QR}\)

∴ QR = \(\frac{6×8}{4}\) = 12 cm

\(\frac{6}{QR}\) = \(\frac{5}{PR}\)

∴ PR = \(\frac{5×8}{4}\) = 10 cm

Hence, the lengths of remaining sides of Δ PQR are QR= 12 cm and PR = 10 cm.

Δ ABC ~ Δ PQR

Then, the corresponding angles are

∠A = ∠P

∠B = ∠Q

∠C = ∠R

Also, the corresponding sides are in proportion.

\(\frac{AB}{PQ}\) = \(\frac{BC}{QR}\) = \(\frac{AC}{PR}\)

\(\frac{10}{5}\) = \(\frac{12}{6}\) = \(\frac{8}{4}\) = \(\frac{2}{1}\)