Question

If D is the midpoint of the hypotenuse AC of a right-angled ∆ABC.prove that BD=½ AC.​

Answer 1

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Answer 2

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$\huge{\underline{\underline{\sf{\orange{</p><p>Question:-}}}}}$

If D is the midpoint of the hypotenuse AC of a right-angled ∆ABC.prove that BD=½ AC.

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$\huge{\underline{\underline{\sf{\orange{</p><p>Solution:-}}}}}$

${\blue{\sf\underline{Given}}}$

A ∆ABC in which ∠B = 90° and D is the midpoint of AC.

${\blue{\sf\underline{To\:Prove}}}$

$bd \: = \frac{1}{2} ac$

${\blue{\sf\underline{Construction}}}$

Produce BD to E such that BD = DE.

Join EC.

${\blue{\sf\underline{Proof}}}$

In ∆ADB and ∆CDE,we have

AD = CD⠀⠀⠀(given),

BD = ED⠀⠀⠀(by construction)

and ∠ADB = ∠CDE

∴∆ADB ≅ ∆CDE ⠀⠀(SAS-criteria)

∴ AB = EC and ∠1 = ∠2 ⠀⠀(c.p.c.t.).

But,∠1 and ∠2 are alternate interior angles.

∴CE || BA.

Now, CE || BA and BC is the transversal.

∴ ∠ABC + ∠BCE = 180° ⠀⠀[∴ ∠ABC = 90°]

⇒ ∠BCE = 90°.

Now,in ∆ABC and ∆ECB,we have

BC = CB ⠀⠀(commom),

AB = EC⠀⠀⠀(proved)

and ∠CBA = ∠BCE ⠀⠀(each equal to 90°).

∴ ∆ABC ≅ ∆ECB ⠀⠀(SAS-criteria).

$\therefore \: AC = EB ⇒ \frac{1}{2} EB = \frac{1}{2} AC ⇒ BD = \frac{1}{2} Ac$

$hence \: BD \: = \frac{1}{2} AC$

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