Let us join AC.
In \(\triangle ABCand\triangle ADC\)
\(\overline { AB } =5cm,\overline { AD } =5cm\quad \Rightarrow \overline { AB } =\overline { AD } \)
\(\overline { BC } =7.5cm,\overline { DC } =7.5cm\quad \Rightarrow \overline { BC } =\overline { DC } \)
\(\overline { AC } =\overline { AC } \)  [common]
\(\therefore \triangle ABC\cong \triangle ADC\)  [SSS congruency]

In the given figure,
DE = IH, EG =FI
and ㄥE = ㄥI
EG = FI
so, EF =Gl
Hence, two sides and one angle of ΔDEF and ΔHIG are equal.
So, ΔDEF ≅ ΔHIG and used congruence criterion is SAS.

We know that, if three sides of one triangle are equal to the three corresponding sides of other triangle, then the triangles are congruent by SSS congruence rule.
Here, ΔART ≅ ΔPEN, by SSS congruence rule
(a) AR = PE
(b) RT = EN
(c) AT = PN

In ΔABE and ΔCDB,
ㄥA = ㄥC = 90°, EB = DB, AE = BC            [given]
i.e. rwo hypotenuse and one side of a right angled ΔABE are respectively equal to the hypotenuse and one
side of other right angled ΔCDB.
So,  ΔABE ≅ ΔCDB  [by RHS congruence rule]

In ΔCAB and ΔDAB, ㄥC = ㄥD = 90°
Hypotenuse AB = Hypotenuse AB     [common]
CA= DB= 2 cm
So, both triangles are congruent by RHS congruence rule.
The correspondence is A ↔️ B, C ↔️ D, B ↔️ A.
In symbolic form, ΔACB ≅ ΔBDA

In the given figure,

AB = XY = 2.5 cm, BC = YZ = 3 cm, AC = XZ = 2 cm Three sides of Δ ABC and corresponding ΔXYZ are equal.
So, ΔABC ≅ ΔXYZ by SSS congruence criterion.

ΔABC and ΔDEF:
We have  

and 
\(\therefore \) The two triangles are congruent (using the ASA congruence rule).
\(\because \) \(A\leftrightarrow F,B\leftrightarrow E and C\leftrightarrow D\)
\(\therefore \) ∆ABC≅∆FED.

Given, ΔABC = ΔPQR
Also given ㄥB = ㄥQ and ㄥC = ㄥR
[from the given figure)
To apply the condition for congruency, included side of one triangle is equal to the included side of the other triangle.
∴ BC=QR
Hence, we use the ASA congruence criterion.

We have, to show that in ΔAMP ≅ ΔAMQ, the missing reasons are as follows:

a = 5

(i) Two line segments are congruent, if they have same length.
(ii) Among two congruent angles, one has a measure of 70°, i.ie measure of the other angle is 70°.
(iii) When we write ㄥA = ㄥB, we actually mean mㄥA = mㄥB.

In the given figure, we have
ㄥA = 90⁰          [right angled at A]
ㄥD = 90⁰           [right angled at D]
∴ ㄥA = ㄥD
BC = BC           [common side]
Also, AC = DB    [given]
So by RHS congruence criterion ΔABC and ΔDCB are congruent to each other.
∴ ΔABC ≅ ΔDCB

Not congruent

In the above figure, AB and CD are two poles whose heights are 9 m and 14m respectively.
\(\Rightarrow\) AB = EC = 9m
and BD = 13m
DE = 14 - 9
= 5m
Now in right ΔBDE, by Pythagoras
BD² = BE² + DE²
13² = BE² + 5²
\(\Rightarrow\) BE² = (13)² - (5)²
= 169-25
BE² = 144
\(\Rightarrow\) BE = \(\sqrt { 144 } \)
\(\Rightarrow\) BE = 12m.
Hence, distance between their feet = 12 m.

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