Free Triangles 02 Practice Test - 9th Grade 

Given, in $△NVY$, NV = VY.


Then, $\angle VYN=\angle YNV$. 
[angles opposite to equal sides of a triangle are equal]
Using angle sum property of a triangle, we get:     
$\angle NVY+\angle VYN+\angle YNV={180}^{\circ }$
So, $\angle NVY+2\angle VYN={180}^{\circ }$.

It is also given that $VY\ne YN$. Hence the options $\angle NVY=\angle VYN$ and $\angle NVY=\angle YNV$ are not correct.

Two triangles can't be congruent if any two sides and one angle of one are equal to any two sides and one angle of the other.
They will be congruent when the angle is included between the equal pair of sides. This is the SAS condition of congruency of triangles.

The SAS congruence rule states:
Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

$\text{In}\Delta AOC\text{and}\Delta AOB,OC=OB\text{(Given)}\angle OBC=\angle OCA\text{(Both angles are right angles)}\text{And}AO\text{is common side}.\text{Hence,}\Delta AOC\cong \Delta AOB\text{(RHS congruence)}\therefore \angle CAO=\angle BAO\text{(CPCT)}\Rightarrow 2x+24=30\Rightarrow x=3\Rightarrow x-3=0$

$\text{In the given figure,}AD=AB\text{(Given)}DC=BC\text{(Given)}AC\text{is common}\therefore \Delta ADC\cong \Delta ABC\text{(SSS congruence)}\text{Hence,}\angle ACD=\angle ACB={30}^{\circ }$

If all the sides of $△$CAT are equal to corresponding sides of $△$RAT, then $△$RAT will be congruent to $△$CAT and the congruence criterion will be SSS.

$\text{Consider}\Delta ABC\text{and}\Delta ADC,AC=AC\text{(Common side of both triangles)}AB=AD\text{(Sides of isosceles triangle)}BC=CD\text{(Median divides a side in two equal parts)}\therefore \Delta ABC\cong \Delta ADC\left(\text{by SSS congruence criterion}\right)\text{So the option}"\Delta ABC\cong \Delta ADC"\text{is correct.}\therefore \angle ACB=\angle ACD\left(CPCT\right)\text{But they are supplementary angles. So,}\angle ACB+\angle ACD={180}^{\circ }\Rightarrow \angle ACB=\angle ACD={90}^{\circ }\therefore AC\text{is perpendicular to}BD\text{and the hence option "AC is perpendicular to BD" is correct.}\text{Also,}\angle BAC=\angle DAC\text{(corresponding angles of congruent triangles)}\text{Hence, the option}"\angle BAC=\angle DAC"\text{is also correct.}\text{Also, area of}\Delta ABC=\text{area of}\Delta ADC=\frac{1}{2}\left(\text{area of}\Delta ABD\right).\Delta ABC\cong \Delta ADC⟹ar\left(\Delta ABC\right)=ar\left(\Delta ADC\right)\text{Hence, the option "Area of}\Delta ABD=\frac{1}{2}\times \text{Area of}\Delta ABC"\text{is also correct.}$

In $△ACB$ and $△CAD$,
$AD=BC$                [Given]
$\angle CAD=\angle ACB$    [Alternate angles]
$CA=AC$                [common side]
$△ACB\cong △CAD$   [SAS congruency]
$\therefore AB=DC$             [CPCT]

In $△$ACD and $△$ABD

$\angle BAD=\angle CAD$
[$\because$ AD is the bisector of $\angle A$]
AD = AD        [common side]
AB = AC         [Given]

$△ACD\cong △ABD$ [SAS congruency]
The triangles are congruent by SAS congruence rule.