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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.