:
Steps: 1 Mark
Proof: 1 Mark
It is given that,
$\mathrm{âˆ}MLN=\mathrm{âˆ}FGH$,
$\mathrm{âˆ}NML=\mathrm{âˆ}HFG$,
$ML=FG.$
$⇒$ The two angles and an included side of one triangle are equal to the corresponding angles and an included side of other triangles.
$∴\mathrm{Δ}LMNâ‰\dots \mathrm{Δ}GFH$ [By ASA congruence criterion]

:
Each Proof: 1 Mark
Steps: 1 Marks
In $\mathrm{Δ}ABC$ and $\mathrm{Δ}ADC$
$AB=DC$ [Given]
$BC=AD$ [Given]
$AC=AC$ [Common side]
$⇒\mathrm{Δ}ABCâ‰\dots \mathrm{Δ}ADC$ [SSS congruency criteria]
$∴\mathrm{âˆ}B=\mathrm{âˆ}D$ [Corresponding parts of congruent triangles]

:
Steps: 1 Mark
Proof: 1 Mark
In $\mathrm{Δ}ADCand\mathrm{Δ}ABC$
$AD=AB$ [Given]
$\mathrm{âˆ}ADC=\mathrm{âˆ}ABC={90}^o$ [Given]
$AC=CA$ [common]
Hence,$\mathrm{Δ}ADCâ‰\dots \mathrm{Δ}ABC$ [By RHS congruence condition]

:
All parts: 2Marks

Given that,
$\mathrm{Δ}PENâ‰\dots \mathrm{Δ}CAR$ and
PN = CR
Corresponding parts of congruent triangle are congruent.
Therefore,thecorresponding sides of congruent triangle are equal.
$⇒PE=CA,EN=AR,PN=CR$.
$⇒$ Also all the corresponding angles of congruent triangles are equal.
$⇒\mathrm{âˆ}P=\mathrm{âˆ}C,\mathrm{âˆ}E=\mathrm{âˆ}A,\mathrm{âˆ}N=\mathrm{âˆ}R$.

:
Properties: 1 Mark
Each proof: 1Mark

In $\mathrm{Δ}ADBand\mathrm{Δ}ADC$
$AB=AC$ [Given]
$\mathrm{âˆ}ADB=\mathrm{âˆ}ADC={90}^{∘}$ [Given]
$AD=AD$ [common]
Hence, $\mathrm{Δ}ADBâ‰\dots \mathrm{Δ}ADC$ [By RHS congruence rule…….(1)]
From (1), $\mathrm{âˆ}BAD=\mathrm{âˆ}CAD$ [Corresponding parts of congruent triangles]
From (1), $BD=DC$ [Corresponding parts of congruent triangles]

:
Proof: 1 Mark
Steps: 2Marks
Consider two triangles, $\mathrm{Δ}ABC$ and$\mathrm{Δ}PQR$ in which,

$\mathrm{âˆ}ABC=\mathrm{âˆ}PQR$
$\mathrm{âˆ}ACB=\mathrm{âˆ}PRQ$
$AB=PQ$
We know that,
$\mathrm{âˆ}ABC+\mathrm{âˆ}ACB+\mathrm{âˆ}BAC={180}^0$
$\mathrm{âˆ}BAC={180}^0−(\mathrm{âˆ}ABC+\mathrm{âˆ}ACB)....\left(i\right)$
Similarly,
$\mathrm{âˆ}QPR={180}^0−(\mathrm{âˆ}PQR+\mathrm{âˆ}PRQ)......\left(ii\right)$
From (i) and (ii),
$\mathrm{âˆ}BAC=\mathrm{âˆ}QPR$
Now, In $\mathrm{Δ}ABC$ and$\mathrm{Δ}PQR$
$\mathrm{âˆ}ABC=\mathrm{âˆ}PQR$ [Given]
$\mathrm{âˆ}BAC=\mathrm{âˆ}QPR$ [ Proved above]
$AB=PQ$[Given]
$⇒\mathrm{Δ}ABCâ‰\dots \mathrm{Δ}PQR$ [ ASA congruency rule]
$⇒$ These triangles are always congruent.

:
Properties: 1 Mark
Proof: 1 Mark
Steps: 2Marks

In a rectangleopposite sides are equal and parallel.
In $\mathrm{Δ}OADand\mathrm{Δ}OCB$,
$\mathrm{âˆ}ODA=\mathrm{âˆ}OBC$
[Alternate interior angles; $ADâˆ\yen BC$ and BDas transversal]
AD = BC [Opposite sides of a rectangle are equal]
$\mathrm{âˆ}OAD=\mathrm{âˆ}OCB$
[Alternate interior angles; $ADâˆ\yen BC$ and ACas transversal]
Hence $\mathrm{Δ}OADâ‰\dots \mathrm{Δ}OCB$ [By ASA congruence rule]
Equating the corresponding parts of congruent triangles, we get:
AO = CO
BO = DO
$⇒$ Diagonals of a rectangle bisect each other.

:
(a) Proof: 2 Marks
(b) Steps: 1 Mark
Final answer: 1 Mark
(a) In $\mathrm{Δ}ABC$ and $\mathrm{Δ}FED$,
$\mathrm{âˆ}B=\mathrm{âˆ}E={90}^{∘}$ [Given]
$\mathrm{âˆ}A=\mathrm{âˆ}F$ [Given]
$BC=ED$ [Given]
$⇒$ Two angles and one side of $\mathrm{Δ}ABC$ are equal totwoangles and one side of$\mathrm{Δ}FED$.
Therefore, $\mathrm{Δ}ABCâ‰\dots \mathrm{Δ}FED$ [AAS congruence rule]
(b)Sum of the angles of a triangle = 180${}^{∘}$
$\mathrm{âˆ}$A + $\mathrm{âˆ}$B + $\mathrm{âˆ}$C = 180${}^{∘}$
$\mathrm{âˆ}$A = 180${}^{∘}$– (50${}^{∘}$ + 60${}^{∘}$) = 180${}^{∘}$ – 110${}^{∘}$ = 70${}^{∘}$

:
Reason: 1 Mark each
(a)If three sides of one triangle are equal to the threesides of the other triangle, then the two triangles are congruent to each other by SSS congruence criterion.
$⇒$ Both triangles look like themirror image of each other.
$⇒$ Both the triangles superimpose on each other.
So, if the sides of a triangleare congruent to the sides of another triangle, the two triangles will be congruent.
(b)Nothing is given or can be said about any of the corresponding sides in this case, As, AAA is not a rule for congruency, the triangles formed may or may not be congruent, depending on if the corresponding parts are equal or not.

:
Application of theorem: 1 Mark
Steps: 2Marks

In $\mathrm{Δ}CATand\mathrm{Δ}R{A}^{′}T$
$CT=RT$ [Given]
$\mathrm{âˆ}CTA=\mathrm{âˆ}RT{A}^{′}$ [Vertically Opposite Angles]
$AT=A^{′}T$ [Given]
$∴\mathrm{Δ}CATâ‰\dots \mathrm{Δ}R{A}^{′}T$ [By SAS congruence rule]
$⇒\mathrm{âˆ}CAT=\mathrm{âˆ}R{A}^{′}T$ [Corresponding parts of congruent triangles]
But $\mathrm{âˆ}CATand\mathrm{âˆ}R{A}^{′}T$ are alternate interior angles.
If the pair of alternate interior angles is equal then the lines are parallel.
$⇒CAâˆ\yen A^{′}R$.