Conditions for the RHS - Right Angle Hypotenuse Side congruence
Two triangles triangle are congruent if the hypotenuse and one side of the one triangle are respectively equal to the hypotenuse and one side of the other.
Experiment to prove Congruence with RHS:
Draw a ∆LMN with ∠M = 90°, LM = 3cm LN = 5 cm,
Also, draw another ∆XYZ with ∠Y = 90°, XY = 3cm and XZ =
We see that ∠M = ∠Y, LM = XY and LN = XZ.
Make a trace copy of ∆XYZ and try to make it cover ∆LMN with X on L, Y on M and Z on N.
We observe that: Two triangles cover each other exactly.
Therefore, ∆LMN ≅ ∆XYZ
Worked-out problems on right angle hypotenuse side congruence triangles (HL postulate):
1. ∆PQR is an isosceles triangle such that PQ = PR, prove that the altitude PO from P on QR bisects PQ.
In the right triangles POQ and POR,
∠POQ = ∠POR = 90°
PQ = PR [since, ∆PQR is an isosceles. Given PQ = PR]
PO = OP [common]
Therefore ∆ POQ ≅ ∆ POR by RHS congruence condition
So, QO = RO (by corresponding parts of congruence triangles)
2. ∆XYZ is an isosceles triangle such that XY = XZ, prove that the altitude XO from X on YZ bisects YZ.
In the right triangles XOY and XOZ,
∠XOY = ∠XOZ = 90°
XY = XZ [since, ∆XYZ is an isosceles. Given XY = XZ]
XO = OX [common]
Therefore ∆ XOY ≅ ∆ XOZ by RHS congruence condition
So, YO = ZO (by corresponding parts of congruence triangles)
3. In the adjoining figure, given that AB = BC, YB = BZ, BA ⊥ XY and BC ⊥ XZ. Prove that XY = XZ
In right triangles YAB and BCZ we get,
YB = BZ [given]
AB = BC [given]
So, by RHS congruence condition
∆ YAB ≅ ∆ BCZ
∠Y = ∠Z (since by corresponding parts of congruence triangles are equal)
XZ = XY (since sides opposite to equal angles are equal)