Contents
Given the hypotenuse and one side of a right-angled triangle.
Construct a right angled triangle given that the hypotenuse is 7 cm and one side is 4 cm.
Two possible triangles can be constructed out of this given data as shown here.
You will find that the two triangles are congruent (check this out).
\(\bigtriangleup{ABC} \cong \bigtriangleup{ZYX}\)
RHS Congruence condition:
If the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and one side of another right-angled triangle, the two triangles are congruent.
RHS Congruence condition Example 1:
If \(\bigtriangleup{ABC}\) is an isosceles triangle such that AB = AC, then altitude AD from A on BC bisects BC.
Solution:
In right triangles ADB and ADC, we have
Hyp.AB = Hyp.AC {Given}
AD = AD {Common side}
So, by RHS criterion of congruence, \(\bigtriangleup{ABD} \cong \bigtriangleup{ACD}\)
=> BD = DC {Since, corresponding parts of above congruent triangles are equal}
RHS Congruence condition Example 2:
AD, BE and CF, the altitudes of \(\bigtriangleup{ABC}\) are equal. Prove that \(\bigtriangleup{ABC}\) is an equilateral triangle.
Solution:
In right triangles BCE and BFC, we have
Hyp.BC = Hyp.BC
BE = CF {Given}
So, by RHS criterion of congruence, we have
\(\bigtriangleup{BCE} \cong \bigtriangleup{BFC}\)
=> \(\angle{B} = \angle{C}\) {Since, corresponding parts of congruent triangles are equal}
=> AC = AB —(1) {Since, sides opposite to equal angles are equal}
Similarly, \(\bigtriangleup{ABD} \cong \bigtriangleup{ABE}\)
=> \(\angle{B} = \angle{A}\) {Since, corresponding parts of congruent triangles are equal}
=> AC = BC —(2) {Since, sides opposite to equal angles are equal}
From (1) and (2), we get
AB = BC = AC
Hence, \(\bigtriangleup{ABC}\) is an equilateral triangle.