ΔABC = ΔABD
[∠ABC = ∠ADB = 90°, ∠ACB = ∠ABD = 90 – X]
∠A (common)
similarly ΔABD ~ ΔBDC
In ΔABC, consider
∠A = X°
∠B = 90°
∠A + C = 90°, ∠C = 90 – X
In ΔABD, ∠ADB = 90°
Therefore ∠A = X°, ∠ABC = 90°
∠A + ∠ABD = 90°
∠ABD = 90 – ∠A = 90 – X;
In ΔBCD, ∠BDC = 90°
∠C = (90 – X); ∠C + ∠DBC = 90°;
∠DBC = 90 – (90 – X) = X°
Hence angles of three triangles are 90°, X° and 90° – X°.
Hence three triangles are similar to each other.