Question

Prove that in a right angle d the median drawn to the hypotenuse is half the hypotenuse in length..​

Answer 1

Answer:

To prove: BD = AC

Construction: Draw DE perpendicular on BC.

Proof: DEL BC (by construction)

and D is the mid-point of hypotenuse AC.

BE = EC

(by converse of mid-point theorem)

Now, A's DBE and DEC

BE = EC

DE DE (common side)

<DEB = <DEC = 90°

(by construction)

..ADBE = ADEC

(by SAS congruency)

→ DC = DB ...(i)

Again, D is mid-point of AC

2DC = AC... (ii)

From (i) and (ii), we get

2DB = AC

→ BD = = AC

Hence proved.

Answer 2

Step-by-step explanation:

Given: ∆ABC is a right angled triangle, right angled at B and BD is its median.

To prove: BD = 1 2 12AC

Construction: Draw DE perpendicular on BC.

Proof: DE ⊥ BC (by construction) and D is the mid-point of hypotenuse AC. ∴BE = EC (by converse of mid-point theorem)

Now, ∆’s DBE and DEC BE = EC DE = DE (common side) ∠DEB = ∠DEC = 90° (by construction) ∴∆DBE ≅ ∆DEC (by SAS congruency)

⇒ DC = DB …(i) Again, D is mid-point of AC 2DC = AC …(ii) From (i) and (ii), we get

⇒ 2DB = AC ⇒ BD = 1 2 12AC

Hence proved.

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