Question

In a right-angled triangle ABC. ∠ABC = 90° and D is the midpoint of AC. Prove that BD = `(1)/(2)"AC"`.

Answer 1

Draw line segment DE || CB, which meets AB at point E.
Now, DE || CB and AB is the transversal,
∴ ∠AED = ∠ABC    ....(corrresponding angles)
∠ABC = 90°             ....(given)
⇒ ∠AED = 90°
Also, as D is the mid-point of AC and DE || CB,
DE bisects side AB,
I.e. AE = BE            ....(i)
In ΔAED and ΔBED,
∠AED = ∠BED      ....(Each 90°)
AE = BE                 ....[From (i)]
DE = DE                ....(Common)
∴ ΔAED ≅ ΔBEd   ....(By SAS Test)
⇒ AD = BD           ....(C.P.C.T.C)
⇒ BD = AC
⇒ BD = `(1)/(2)"AC"`.