In \(\triangle \)ABD and \(\triangle \)ACD, we have \(\angle ADB=\angle ADC\)  \([\because AD\bot BC]\)
side \(\overline { AD } \)=side \(\overline { AD } \)   [common]
Hypt.\(\overline { AB } \)=Hypt.\(\overline { AC } \)    [Given]
:. Using RHS congruency, we get \(\triangle \)ABD \(\cong \)\(\triangle \)ACD.
\(\Rightarrow \)Their corresponding parts are equal 
\(\therefore \) BD=CD
\(\Rightarrow \)D is the mid-point of BC

Given: A triangle ABC
such that AB = AC.
Bisector of ㄥA is AD.
To prove: ㄥADB = ㄥADC=90° and ​​​​​​​\(\overline { BD } =\overline { CD } \)
Proof: In \(\triangle \)ADB and \(\triangle \)ADC, We have
\(\overline { AD } =\overline { AD } \)
ㄥBAD=ㄥCAD
[\(\because \) AD is the bisector of ㄥBAC]
\(\overline { AB } =\overline { AC } \) [Given]
\(\therefore \)\(\triangle \)ADB ≅\(\triangle \)ADC [SAS congruency]
\(\Rightarrow \) Their corresponding parts are equal.
\(\therefore \quad \overline { BD } =\overline { CD } \)
i.e. the side BC is bisected at D.
Also ㄥADB =ㄥADC
But ADB and ADC form a linear pair.
\(\therefore \)ㄥADB +ㄥADC=180°
ㄥADB+ㄥADB=180°
\(\Rightarrow \) 2ㄥADB=180°
\(\Rightarrow \)\(\angle ADB=\frac { 180^{ \circ } }{ 2 } \Rightarrow \angle ADB={ 90 }^{ \circ }\)
i.e. AD \(\bot \)BC
Hence AD is the perpendicular bisector of BC.
​​​​​​​

Given: A ΔABC in which \(\overline { AB } =\overline { AC. } \)
To prove:\(\angle B=\angle C\)
Construction: Draw AD 丄 BC
Proof: In right ΔADB and right ΔADC, we have
side AD = side AD [Common]
Hypt. AB = Hypt. AC [Given]
⇒ ΔADB ≅ΔADC [RHS congruency]
\(\therefore \)Their corresponding parts are equal.
⇒ ㄥB=ㄥC

In the given triangles.
ΔABC and ΔQPR
AB = QP= 3·5
AC= QR = 5·1
BC= PR = 7·1
So there exists S.S.S. congruency.
Hence, ΔABC ≌ ΔQPR

Here, according to the question, the figure is as follows

From figure, we see that
\(\overline { RP } = \overline { QP } ,\overline { RS } =\overline { QS } \)
and \(\overline { SP }=\overline { SP } \)     [common]
So, by SSS congruence criterion ΔPSQ and ΔPSR are congrunet to each there.

In ΔABC and ΔPQR, we have
AB = 3.5 cm and PQ = 3.5 cm, so AB = PQ
AC = 3.5 cm and PR = 3.5 cm, so AC = PR
LA=60° and LP=60°, so ㄥA = ㄥP
Hence, two sides and an angle in ΔABC and ΔPOR are equal. Therefore, the two triangles are congruent.
∴ ΔABC ≅ ΔPOR

In the following figure, clearly, the side included between ㄥA and ㄥB is \(\bar { AB } \).

If  ΔPOR ≅ ΔBCA, then corresponding congruent parts of ΔBCA \(\bar { QR } \leftrightarrow \bar { CA } \).

If  ΔPOR ≅ ΔBCA, then corresponding congruent parts of ΔBCA ㄥQ ↔️ ㄥC

All the corresponding congruent parts of  ΔABC and ΔPOR are
\(\angle A ↔️ \angle P, \angle B ↔️ \angle Q, \angle C ↔️ \angle R\) , and \(\overline { AB } \leftrightarrow \overline { PQ } ,\overline { BC } \leftrightarrow \overline { QR } ,\overline { CA } \leftrightarrow \overline { RP } \).

We know that, if two sides and the angle included between them of a triangle are equal to two corresponding sides and the angle included them of other triangle, then the triangles are congruent by SAS
congruence rule.
Here, ΔART ≅  ΔPEN and ㄥT = ㄥN
So, by SAS congruence rule,
(a) RT = EN
(b) PN = AT 

In ΔPQS and ΔPRS, ㄥPSQ = ㄥPSR = 90°
Hypotenuse PQ = Hypotenuse PR = 3 cm
PS = PS                  [common]
Therefore, by RHS congruence rule, two triangles are congruent. The correspondence is P ↔️ P, Q ↔️ R,
S ↔️ S.
In symbolic form, ΔPQS ≌ ΔPRS

In ΔPQR and ΔMNL,
ㄥR = ㄥL = 60° [given]
RQ= LN = 6 cm [given]
ㄥQ = ㄥN = 30° [given]
Therefore, by ASA congruence rule, two triangles are congruent.
The correspondence is P ↔️ M, Q ↔️ N and R ↔️ L.
In symbolic form, ΔPQR ≅ ΔMNL

ㄥE ↔️ ㄥQ

If ΔDEF ≅ ΔPQR, then corresponding congruent part(s) of the triangle is ㄥF ↔️ ㄥR