Question

If D is the midpoint of the hypotenuse AC of a right angled triangle ABC, prove that BF = ½AC​

Answer 1

I think ..it should be correction there ...it would be BD = 1/2 AC in a right angle triangle

Answer 2

$\bf\underline{\underline{\pink{Question:-}}}$

★If D is the midpoint of the hypotenuse AC of a right angled triangle ABC, prove that BF = ½AC

$\bf\underline{\underline{\blue{Given:-}}}$

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

$\bf\underline{\underline{\red{To Prove:-}}}$

★BD = ½ of AC

$\bf\underline{\underline{\green{Construction:-}}}$

★Produce BD to E such that BD = DE. Join EC

$\bf\underline{\underline{\orange{Proof:-}}}$

In ∆ADB and ∆CDE, we have

AB = CD (given)

BD = ED (by construction)

and ∠ADB = ∠CDE ( vert. opp. angle)

∴ ∆ADB ≅ ∆CDE (S.A.S-criteria)

∴ AB = EC and ∠1 = ∠2 (c.p.c.t)

But ∠1 and ∠2 are alternate interior angles.

∴ ∠ABC + ∠BCE = 180° [ co-interior angle]

==> 90° + ∠BCE = 180° [∵ ∠ABC = 90°]

==> ∠BCE = 180° – 90°

==> ∠BCE = 90°

Now, in ∆ABC and ∆ECB, we have

BC = CB (given)

AE = EC (proved)

and ∠CBA = ∠BCE

∴ ∆ABC ≅ ∆ECB (S.A.A-criteria)

∴ AC = EB => ½EB = ½AC => BD = ½AC

Hence, BD = ½AC