(i) Given, ray AZ i.e. AC is the bisector of ㄥDAB as well as ㄥDCB.
∴ ㄥDAC = ㄥBAC and ㄥDCA = ㄥBCA
Now, three pairs of equal parts in ΔBAC and ΔDAC are
ㄥDAC = ㄥBAC
[since, AC is the bisector of ㄥDAB]
AC = AC [common]
and ㄥDCA = ㄥBCA [since,AC is the bisector of ㄥDAB]
(ii) Yes, in ΔBAC and ΔDAC, we have
ㄥDAC = ㄥBAC, AC = AC, ㄥDCA = ㄥBCA
So, by ASA congruence rule, two triangles are congruent.
The correspondence is A ↔️ A, C ↔️ C and D ↔️ B.
In symbolic form, ΔBAC  ≅ ΔWAC
iii) Yes, here ΔBAC ≅ ΔDAC
We know that, the corresponding parts of two congruent triangles are equal.
So, AB = AD [corresponding sides]
(iv) Yes, Here, ΔBAC ≅ ΔDAC
We know that, the corresponding parts of two triangles are equal.
So, CD = CB [corresponding sides]

If ΔABC and ΔXYZ are congruent, then following should be true

AB=XZ     [4.8 cm each]
ㄥA = ㄥX      [90° each]
AC = XY [6.8 cm each]
These are not congruent because it is not follows the RHS congruence rule perfectly.
ΔABC ≇ ΔXYZ

In ΔPQR and ΔSTU,

PQ = ST       [3.5 cm each]
QR = TU       [4 cm each]
ㄥQ = ㄥT      [60° each]
So, by SAS congruence rule, two triangles ΔPQR and ΔSTU are congruent.
The correspondence is P ↔️ S, Q ↔️ T and R ↔️ U. In symbolic form, ΔPQR ≅ ΔSTU.

According to the question, we have following figure:

Quadrilateral ABCD can be divided into two triangles i.e. ΔABD and ΔBDC.
In two triangles,
AD ≠ CD
AB ≠ BC
BD = BD          [common side]
So, ΔABD ≅ ΔBCD are not true.

In the given figure, we have
BD = CE, EC 丄 AB and AB 丄 AC,
So, ㄥBEC = 90⁰  ⇒ ㄥBDC = 90⁰
So, ㄥBEC = ㄥBDC
and BC = BC
[common side is hypotenuse of right angled triangle]
Since, two sides and one angle of ΔBDC and ΔBEC are equal. So, ΔBCD ≌ ΔCBE
[by RHS congruence criterion]

Congruent

Congruent

2x+3y=220°

ㄥAOC=ㄥBOD [Vertically opposite angles]
We have two triangles AOC and BOD
Such that AC = BD (Given)

\(\because \)Vertically opposite angles are equal
\(\therefore \)ㄥAOC=ㄥBOD      [Each angle is 35°]               
Now, using the angle sum property of a triangle, we have
ㄥA of MOC = 180° - (75° + 35°)
= 180° - 110° = 70°
and ㄥBof ΔBOD=180° - (75° + 35°)
= 180° - 110° = 70°
i.e.ㄥA=ㄥB
Now, side AC is between ㄥA and ㄥC and side BD is between ㄥB and ㄥD
\(\therefore \) By ASA congruence rule, ΔAOC ≅ ΔBOD
Note: If two angles of a triangle are known, then we can find the third angle of the triangle. Thus, in case of two angles of a triangle being equal to two corresponding angles of another triangle, then the third angles of both the triangles are equal.

Let AB is the street and C be the foot of ladder. Let D and E be windows at heights of 8 m and 15 m respectively from the ground.

Then CD and CE are the positions of ladder. From the right angle ΔDAC, By Pythagoras Theorem,
CD² = AC² + AD²
⇒ AC² = CD² - AD²
= 17²-8²
= 289-64
= 225
⇒AC = \(\sqrt { 225 } \) = 15
Again from right ΔEBC,by Pythagoras Theorem,
CE² = BC² + BE²
BC² = CE² - BE²
= 17²-15²
= 289-225
= 64
\(\Rightarrow\) BC = \(\sqrt { 64 } \)
= 8m
∴ Width of street,
AB = AC + BC
= 15 + 8 = 23 m