It is true....becoz in dis following conditions r possible......1) SAS when prependicular n basebr equal2) RHS. if hypotnuse n base/ prependicular r equal... as we have one angle (90) equal these two triangles r congruent

Explanation:

The statement is true and is known as the Hypotenuse-Leg (HL) theorem.

HL Theorem:
If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the two triangles are congruent.

Proof:
Let ΔABC and ΔDEF be two right triangles such that AB = DE, BC = EF and ∠B = ∠E = 90°.

We need to prove that ΔABC ≅ ΔDEF.

From the given information, we have:

AB = DE (Given)
BC = EF (Given)
∠B = ∠E = 90° (Given)

We know that in a right triangle, the hypotenuse is the longest side. Therefore, we can say that AC > AB and DF > DE.

Let's consider two cases:

Case 1: AC > DF
In this case, we draw a line segment DG perpendicular to EF.

Now, we have two right triangles ΔDFG and ΔACB as shown below:



We can see that:

∠DGF = ∠CAB (Both are 90°)
∠DFG = ∠ACB (Vertically opposite angles)
DG = AB (Perpendicular from D to EF bisects EF)
DF = AC (Given)

Therefore, by Angle-Side-Angle (ASA) congruence criterion, we can say that ΔDFG ≅ ΔACB.

Now, we can say that:

∠FDE = ∠GDC (Alternate Interior Angles)
∠DEF = ∠DGF + ∠FDE = ∠CAB + ∠GDC = ∠GED (Angles on a Straight Line)
DE = DG (Perpendicular from D to EF bisects EF)
DF = AC (Given)

Therefore, by Angle-Side-Angle (ASA) congruence criterion, we can say that ΔDEF ≅ ΔDGF.

By transitive property of congruence, we can say that ΔABC ≅ ΔDEF.

Case 2: AC < />
In this case, we draw a line segment CH perpendicular to AB.

Now, we have two right triangles ΔACH and ΔDEF as shown below:



We can see that:

∠ACH = ∠DEF (Both are 90°)
∠CAH = ∠FDE (Given)
CH = EF (Perpendicular from E to AB bisects AB)
AC = DE (Given)

Therefore, by Angle-Side-Angle (ASA) congruence criterion, we can say that ΔACH ≅ ΔDEF.

Now, we can say that:

∠BAC = ∠EDF (Given)
∠ABC = ∠DEF = 90° (Both are right angles)
AB = DE (Given)

Therefore, by Angle-Side-Angle (ASA) congruence criterion, we can say that ΔABC ≅ ΔDEF.

Conclusion:
Hence, we have proved that if any two sides of a right triangle are respectively equal to two sides of other