Question

In the given figure, ΔABC is right angled at B. ACDE and BCGF are squares.

Prove that

1. ΔBCD ≅ ΔACG
2. AG = BD

Answer 1

Given:

ΔABC is right-angled at B.

ACDE and BCGF are squares.

To Prove:

1. ΔBCD ≅ ΔACG
2. AG = BD

Proof:

i. Proving ΔBCD ≅ ΔACG

1. In square ACDE, AC = CD (Sides of square are equal).

2. In square BCGF, BC = CG (Sides of square are equal).

3. Given ΔABC is right-angled at B, ∠ABC = 90°.

4. Since ACDE and BCGF are squares, ∠ACD and ∠BCG are also right angles (90° each).

5. Consider ΔBCD and ΔACG:

BC = AC   ...(Both are sides of the right triangle ΔABC where AC is hypotenuse and BC is one of the legs)

CD = CG   ...(Sides of squares ACDE and BCGF)

∠BCD = ∠ACG = 90°   ...(Angles of the squares)

6. By RHS (Right angle-Hypotenuse-Side) congruence rule, ΔBCD ≅ ΔACG.

ii. Proving AG = BD

1. From congruent triangles ΔBCD and ΔACG, corresponding parts are equal (CPCT).

2. So, AG = BD.

Thus, the required proofs are complete:

ΔBCD ≅ ΔACG

AG = BD