How to check for similarity in triangles? - class-IX

$SAS$ is a congruency condition not a smililarity condition of triangle

According to the Converse of the Basic Proportionality Theorem, if PM/PQ = PN/PR, then MN || QR. Here, PM/PQ = 0.18/1.28 = 18/128 = 9/64. PN/PR = 0.36/2.56 = 36/256 = 9/64. Since the ratios are equal, the lines are parallel.

In $\triangle$s, $OAB$ and $OCD$
$\dfrac{OC}{OA} = \dfrac{OD}{OB}$ (Given)
$\angle AOB = \angle COD$ (Vertically opposite angles)
Thus, $\triangle OAB \sim \triangle OCD$ (SAS rule)

In $\triangle ABC$, $M$ is mid point of $AB$ and $N$ is mid point of $AC$
$D$ is any point of BC
Now, Join AD and MN such that they met at O
In $\triangle ABC$
M is mid point of AB and N is mid point point of AC
Hence, $MN \parallel BC$ and $MN = \frac{1}{2} BC$

Now, In $\triangle ABD$
$MO \parallel BC$ and M is mid point of AB
Thus, $O$ is mid point of AD
Hence, $MN$ bisects $AD$

Given: $D$ is mid point of $AB$ and $E$ is mid point of $BC$, $F$ is any point on $AC$ and $EF \parallel AB$

Now, in $\triangle ABC$,
E is mid point of BC and $EF \parallel AB$
By Mid point Theorem, $F$ is mid point of $AC$

Also, D is mid point of AB and F is mid point of AC
Hence, by mid point theorem, $DF \parallel BE$
Since, $DF \parallel BE$ and $EF \parallel AB or BD$
Hence, BEFD is parallelogram.

Given triangles DEF and PQR, with angle D = angle Q and angle R = angle E. By AA similarity, triangle DEF is similar to triangle QRP. The corresponding sides are proportional: DE/QR = EF/RP = DF/QP. Option B claims DE/PQ = EF/RP, which is not necessarily true because PQ is not the corresponding side to DE.

Given triangle ABC and DEF. Angle A = Angle E = 40 degrees. AB/ED = AC/EF. This satisfies the SAS similarity criterion, so triangle ABC is similar to triangle EDF. Therefore, Angle B = Angle D and Angle C = Angle F. We are given Angle F = 65 degrees, so Angle C = 65 degrees. In triangle ABC, Angle A + Angle B + Angle C = 180. 40 + Angle B + 65 = 180 => Angle B = 180 - 105 = 75 degrees.

Given: D is mid point of BC. $DM \perp AB$ and $DN \perp AC$, $DM = DN$
Now, In $\triangle DMB$ and $\triangle DNC$,
$DM = DN$ (Given)
$\angle DMB = \angle DNC$ (Each $90^{\circ}$)
$BD = DC$ (D is mid point of BC)
Thus, $\triangle DMB \cong \triangle DNC$ (SAS rule)
Thus, $\angle B = \angle C$ (By cpct)
hence, $\triangle ABC$ is an Isosceles triangle.

Equilateral triangles are similar triangles.
In similar triangles, the ratio of their corresponding sides is the same as the ratio of their medians.
Hence, ratio of sides = $3: 2$

The area of two similar triangles are proportional to the square of their corresponding sides

If two triangles are equals than the ratio of their square is equal to the ratio of their corresponding sides.

If two triangles are similar ,then the ratio of their corresponding sides and altitude are also same. Therefore the ratio of the altitude is 2:3.

If two triangles are equal then their corresponding angle are also equal

Since triangle ABC and DEF are similar 

If two triangles are similar then the ratio of their perimeter is equal to the ratio of their corresponding sides.

Since triangle ABC anD PQR are similar 

$\because \triangle LMN=\triangle PQR$ then

If two triangles are equals than the ratio of their square is equal to the ratio of their corresponding sides.

If two triangles are same than their sides and corresponding attitudes are also same then the ratio of the corresponding altitude is 2:3. 

$P _1=AB+BC+AC=60  cm$

If two sides and the included angle of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.
Therefore, D is the correct answer.

If the corresponding angles are equal then the triangles are similar by $AAA$ similarity criteria.

$\Delta {ABC} \sim \Delta PQR, \angle{B} = \angle{Q}$ is said to be SAS similarity of postulate.
Because, SAS Similarity Postulate states, "If an angle of one triangle is congruent to the corresponding angle of another triangle and the sides that include this angle are proportional, then the two triangles are similar."

As we consider only sides, therefore, SSS similarity is used.
Option A is correct.  

For $\triangle ABC $ and $\triangle PQR$,
$m\angle A = m\angle R$
$m\angle C= m\angle Q$
$\therefore $ by AA criteria for similarity 

If two angles of a triangle is equal to two angles of another triangle, then the third angle of both triangles will be equal.
$\therefore$By AAA Theorem of Similarity, the two triangles are similar.

$\triangle ABC \sim \triangle BDE$                            (both are equilateral triangles)

By the Basic Proportionality Theorem converse, if AD/AB = AE/AC = 1/4, then DE is parallel to BC and triangle ADE is similar to triangle ABC with a scale factor of 1/4. Therefore, DE = (1/4)BC, not 1/8.

In $\triangle$ APB and $\triangle$ CPD,
$\angle APB = \angle CPD$ (Vertically opposite angles)
$\angle ABP = \angle CDP$ (Alternate angles of parallel sides AB and CD)
$\angle BAP = \angle DCP$ (Alternate angles of parallel sides AB and CD)
Hence, $\triangle APB \sim \triangle CPD$ (AAA rule)

In $\triangle$s, APB and ECB,

In $\triangle$s, APB and ECB,
$\angle ABP = \angle EBC $ (Common angle)
$\angle PAB = \angle CEB$ (Corresponding angles of parallel lines)
$\angle APB = \angle ECB $ (Third angle of the triangle)
Thus $\triangle APB \sim \triangle ECB$ 
Hence, $\frac{AB}{EB} = \frac{BP}{BC}$ (Corresponding sides of similar triangles)
$2 = \frac{BP}{BC}$
$BP = 2 BC$
$BP = 2 AD$  (BC = AD)

Given: $AO = 2 CO$ or $\dfrac{AO}{CO} = 2$
Also given, $BO = 2 DO$ or $\dfrac{BO}{DO} = 2$
In $\triangle AOB$ and $\triangle COD$, we know 
$\angle AOB = \angle COD$
$\dfrac{AO}{CO} = \dfrac{BO}{DO}$
Thus, $\triangle AOB \sim \triangle COD$ (SAS rule)

Given: $AB = AC$, $PQ \perp AB$ and $PR \perp AC$
Since, $AB = AC$
$\angle ABC = \angle ACB$...(I) (Isosceles triangle property)

Now, In $\triangle PBQ$ and $\triangle PRC$
$\angle PBQ = \angle PCR$ (From I)
$\angle PQB = \angle PRC$ (Each $90^{\circ}$)
$\angle QPB = \angle RPC$ (Third angle)
Thus, $\triangle QPB \sim \triangle RPC$ (AAA rule)
Hence, $\dfrac{PQ}{PR} = \dfrac{PB}{PC}$
$\dfrac{15}{9} = \dfrac{PB}{12}$
$PB = \dfrac{15 \times 12}{9}$
$PB = 20$ cm