Right Angle Triangle Theorem

Triangle is a polygon which has three sides and three vertices. Triangles having same shape and size are said to be congruent. Similarity of triangles uses the concept of similar shape and finds great applications. Triangles are said to be similar if:

a. Their corresponding angles are equal.

b. Their corresponding sides are in the same ratio.

Theorem:In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Similarity Of Triangles

Proof: We have a \(Δ ABC\) in which \(AC^2\) = \(AB^2 ~+~ BC^2\)

We need to prove that \(\angle{B}\) = \(90^{\circ}\)

In order to prove the above, we construct a triangle \(PQR\) which is right angled at \(Q\) such that \(PQ\)

= \(AB\) and \(QR\) = \(BC\).

From triangle \(PQR\), we have

\(PR^2\) = \(PQ^2 ~+~ QR^2\) (According to Pythagoras theorem,as \(\angle{Q}\) = \(90^{\circ}\))

or, \(PR^2\) = \(AB^2~ + ~BC^2\) (By construction) …… (1)

We know that \(AC^2\) = \(AB^2 ~+~ BC^2\) (Which is given) …………(2)

So, \(AC\) = \(PR\) [From equation (1) and (2)]

Now, in \(Δ ABC\) and \(Δ PQR\),

\(AB\) = \(PQ\) (By construction)

\(BC\) = \(QR\) (By construction)

\(AC\) = \(PR\) [Proved above]

So, \(Δ ABC~ ≅ ~Δ PQR\) (By SSS congruence)

Therefore, \(∠B\) = \(∠Q\) (CPCT)

But \(∠Q\) = \(90^{\circ}\) (By construction)

So, \(∠B\) = \(90^{\circ}\)

Hence the theorem is proved.

Illustration 1: The angle \(\angle{PRQ}\) = \(90^{\circ}\) and \(RS\) is perpendicular to \(PQ\). Prove that \(\frac{QR^2}{PR^2}\) = \(\frac{QS}{PS}\) .

Similarity Of Triangles

Solution: We can see that \(\triangle~PSR\) ~ \(\triangle~PRQ\)

According to the property of similar triangles we have:

\(\frac{PR}{PQ}\) = \(\frac{PS}{PR}\)

or it can also be written as, \(PR^2\) = \(PQ.PS\) ……….(1)

Similarly we have \(\triangle~QSR\) ~ \( QRP\)

So we have,\(\frac{QS}{QR}\) = \(\frac{QR}{QP}\) Or it can also be written as, \(QR^2\) = \(QP.QS\) ……….. (2)

By dividing the equations (2) by (1) we can deduce that:

\(\frac{QR^2}{PR^2}\) = \(\frac{QP .QS}{PQ.PS}\) = \(\frac{QS}{PS}\)<

Hence the theorem is proved.

This article covers the theorem on similarity of triangles. For any further information on this topic install Byju’s the learning app.

Leave a Comment

Your email address will not be published. Required fields are marked *