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