AAA Criterion Of Similarity Of Triangles

We have various criteria to prove two triangles similar. AAA criterion is one of these.

We can check the similarity of any two triangles using AAA criterion of similarity if any two angles of each triangle are given so, AAA criterion is same as AA criterion.

AA criterion "If two triangles are equiangular, then their corresponding sides are proportional." can be proved as below.

Given: ΔABC and ΔPQR where ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R.

https://img-nm.mnimgs.com/img/study_content/editlive_lp/33/2013_05_29_12_55_45/1.png

To prove:https://img-nm.mnimgs.com/img/study_content/editlive_lp/33/2013_05_29_10_48_38/mathmlequation2764178644931880778.png

Construction: Mark X and Y on AB and AC respectively such that AX = PQ and AY = PR.

Proof:

In ΔAXY and ΔPQR,

AX = PQ [By construction]

∠A = ∠P [Given]

AY = PR [By construction]

So, by SAS postulate, ΔAXY ≡ ΔPQR.

[Note: The symbol '≡' stands for congruency]

⇒XY = QR and ∠X = ∠Q [CPCT]

Now, ∠X = ∠B [∠X = ∠Q = ∠B]

∴XY||BC [∠X and ∠B are corresponding angles]

https://img-nm.mnimgs.com/img/study_content/editlive_lp/33/2013_05_29_12_57_44/mathmlequation6904500104195715890.png

Hence, AA criterion is proved.

Now, look at the following triangles.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/444/946/10.1.6.3.1_ok_SS_html_m3f33319.png

Here, ∠B = ∠E = 50°

and ∠C = ∠ F = 40°

Then, using AAA similarity criterion, ΔABC is similar to ΔDEF.

In symbolic form, we can write ΔABC ∼ ΔDEF. In symbolic form, the order of vertices is very important. For the above triangles, we cannot write ΔABC ∼ ΔEFD because ∠B = ∠E and ∠C = ∠F

Let us look at one more application of the AAA similarity criterion in the video in order to get a better understanding of this concept.

Converse of AAA criterion is also true which states that:

If two triangles are similar then their corresponding angles are equal.

For example, if ΔABC ∼ ΔDEF then ∠A = ∠D, ∠B = ∠E and ∠C = ∠ F.

Note: In some state boards, the symbol "|||"is used for similarity.

I.e., ΔABC ∼ ΔDEF may also be written as ΔABC ||| ΔDEF.

Let us now look at some more problems based on AAA similarity criterion.

Example 1: In the following figure, if DE || BC, then prove the following.

(a) ΔABC ∠ ΔADE

(b) ΔDFE ∠ ΔCFB

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/444/946/10.1.6.3.1_ok_SS_html_62d568eb.jpg

Solution:

(a) In ΔABC and ΔADE,

∠BAC = ∠DAE (Common to both)

∠ADE = ∠ABC (Since DE is parallel to BC, ∠ADE and ∠ABC are corresponding angles) By AAA similarity criterion,

ΔABC ∼ ΔADE

(b) In ΔDFE and ΔBFC,

∠DFE = ∠BFC (Vertically opposite angles)

∠EDF = ∠BCF (Alternate angles) By AAA similarity criterion,

ΔDFE ∼ ΔCFB

Example 2: In the given figure, if WY || ZX, then prove that ΔOWY ∠ ΔOXZ.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/444/946/10.1.6.3.1_ok_SS_html_7ec9d021.png

Solution:

Here, WY || ZX

Now, in ΔOWY and ΔOZX,

∠WOY = ∠ZOX (Vertically opposite angles)

∠OWY = ∠OXZ (Alternate angles)

∠OYW = ∠OZX (Alternate angles)

By AAA similarity criterion of triangles, ΔOWY ∼ ΔOXZ

SSS Criterion of Similarity of Triangles

Now that we have understood the concept of AAA similarity criterion, we will try and understand another similarity criterion which is the SSS similarity criterion. It involves the ratio of the corresponding sides of the two triangles.

Converse of SSS criterion is also true which states that:

If two triangles are similar then their corresponding sides are proportional.

For example, if ΔABC ∼ ΔPQR then https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_05_08_13_06_18/mathmlequation748259041717846278.png.

Let us solve some problems to understand this concept better.

Example 1: If PQR is an isosceles triangle with PQ = PR and A is the mid-point of side QR, then prove that ΔPAQ is similar to ΔPAR.

Solution:

It is given that ΔPQR is an isosceles triangle and PQ = PR.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/445/484/10.1.6.3.2_ok_html_1156a464.png

In triangles PAQ and PAR, PQ = PR

Also, A is the mid-point of QR, therefore

QA = AR

And, PA = PA (Common to both triangles)

Therefore, we can say that

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/445/484/10.1.6.3.2_ok_html_6ff49014.gif

∴ Using SSS similarity criterion, we obtain ΔPAQ ∼ ΔPAR

Example 2: In the following figure, E and D are the mid-points of the sides BC and AC respectively. Prove that ΔABC∠ΔDEC.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/445/484/10.1.6.3.2_ok_html_59e18fd8.png

Solution:

It is given that E is the mid-point of BC.

∴ BE = EC

Now, BC = BE + EC

⇒ BC = 2EC

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/445/484/10.1.6.3.2_ok_html_m2b9ea108.gif

Similarly, D is the mid-point of AC, therefore

AC = 2DC

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/445/484/10.1.6.3.2_ok_html_m7bbf73d1.gif

Also, from the figure,

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/445/484/10.1.6.3.2_ok_html_401d2669.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/445/484/10.1.6.3.2_ok_html_m3c976646.gif

By SSS criterion of similarity of triangles,

ΔABC ∼ ΔDEC

Example 3: In the following figure, the lines XC and YC of same length are drawn such that C is the mid-point of AB. If AX = BY, then find the measure of the following angles.

[1] ∠BYC (c) ∠CAX

[2] ∠CBY (d) ∠ACX

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/445/484/10.1.6.3.2_ok_html_m730c46d0.png

Solution:

In the triangles CAX and CBY,

CX = CY (Given)

CA = CB (C is the mid-point of AB)

AX = BY (Given)

Therefore, by SSS similarity criterion, ΔCAX ∼ ΔCBY

We know that the corresponding angles of similar triangles are equal.

∴ ∠AXC = ∠BYC = 40°

⇒ ∠BYC = 40°

Also, ∠ACX = ∠BCY

Lethttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/445/484/10.1.6.3.2_ok_html_bc0d09a.gif

Therefore, x + x + 120° = 180° (∠ACX, ∠BCY, and ∠XCY form a linear pair)

⇒ 2x = 180° − 120° = 60°

x = 30°

∴ ∠ACX = ∠BCY = 30°

Now, by angle sum property in ΔACX, we obtain

30° + ∠CAX + 40° = 180°

⇒ ∠CAX = 180° − 70° = 110°

∴ ∠CBY = ∠CAX = 110°

Thus, we obtain

1. ∠BYC = 40°

2. ∠CBY = 110°

3. ∠CAX = 110°

4. ∠ACX = 30°

Example 4: ABCD is a square and PQS is an isosceles triangle with PQ = PS and R is the mid-point of QS. If ΔABD ∠ ΔRPQ, then prove that ΔCBD ∠ ΔRPS.

Solution:

ABCD is a square and PQS is an isosceles triangle.

Therefore, AB = BC = CD = DA

And, PQ = PS

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/446/486/10.1.6.3.3_SMT_PIY_NVN_KSB_SS_html_254bd680.png

It is also given that ΔABD ∼ ΔQRP.

Inhttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/446/486/10.1.6.3.3_SMT_PIY_NVN_KSB_SS_html_m3ac55ec1.gif

AB = CB (Sides of a square)

BD = BD (Common side)

DA = DC (Sides of a square)

By SSS similarity criterion,

ΔABD ∼ ΔCBD … (2)

Now, in ΔRPQ and ΔRPS,

RP = RP (Common side)

PQ = PS (Equal sides of an isosceles triangle)

QR = SR (R is the mid-point of QS)

Therefore, ΔRPQ ∼ ΔRPS … (3)

However, ΔABD ∼ ΔRPQ

Therefore, from (2) and (3), we obtain

ΔCBD ∼ ΔRPS

SAS Criterion Of Similarity Of Triangles

Look at the following figures.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/446/486/10.1.6.3.3_SMT_PIY_NVN_KSB_SS_html_m367dcb8a.png

Is there any similarity between them?

We can see that in both the triangles, the lengths of two sides are given and also the measure of the included angle is given. Now, let us compare the sides of the triangles and observe the result we obtain.

On taking the ratio of the sides, we obtain

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/446/486/10.1.6.3.3_SMT_PIY_NVN_KSB_SS_html_m6f1ed6d2.gif

Therefore, we observe that the sides of the triangles are in the same ratio i.e., we can say that the sides of the triangles are proportional.

Using the above fact, can we say that the given triangles are similar?

To know the answer, let us first know about a similarity criterion known as SAS similarity criterion.

SAS similarity criterion can be stated as follows.

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar”.

Using this criterion, we can check the similarity of any two triangles, if the two sides and the included angle between them are given.

In the above example, ∠A = ∠D = 65° and the sides including these angles are in the same

proportion i.e., . https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/446/486/10.1.6.3.3_SMT_PIY_NVN_KSB_SS_html_79f5f82c.gifThus, we can say that ΔABC is similar to ΔDEF.

In symbolic form, we can write ΔABC ∼ ΔDEF. For writing the symbolic form, the order of the vertices is very important.

For example, consider the following figure.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/446/486/10.1.6.3.3_SMT_PIY_NVN_KSB_SS_html_6bb4c8dd.png

Here, ΔABC and ΔDEF are similar triangles as two sides of both the triangles are proportional and the angles included between them are also equal.

Therefore, we can write ΔABC ∼ ΔEFD.

Let us now look at some more examples to understand this concept better.

Example1: If PQRS is a parallelogram, then prove that ΔSOR is similar to ΔPOQ.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/446/486/10.1.6.3.3_SMT_PIY_NVN_KSB_SS_html_m7d442a2c.png

Solution:

Consider ΔSOR and ΔPOQ.

Since PQRS is a parallelogram, the diagonals bisect each other.

∴ SO = OQ and PO = OR

and ∠POQ = ∠SOR (Vertically opposite angles)

By SAS similarity criterion, we obtain

ΔSOR ∼ ΔQOP

Example2: ΔABC is an isosceles triangle with AB and AC as the equal sides. The points D and E divide the side BC into three equal parts as shown in the figure. Prove that ΔABD ∠ ΔACE.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/133/172/446/486/10.1.6.3.3_SMT_PIY_NVN_KSB_SS_html_mc956b52.png

Solution:

Since ABC is an isosceles triangle,

AB = AC

∠ABC = ∠ACB (Angles opposite to equal sides are equal in an isosceles triangle)

It is given that the points D and E divide the side BC in three equal parts. Therefore,

BD = DE = EC

In ΔABD and ΔAEC,

AB = AC

BD = EC

∠ABD = ∠ACE

By SAS similarity criterion,

ΔABD ∼ ΔACE