Question

If D is the midpoint of the hypotenuse AC of a right angled triangle ABC,

prove that BD = 1/2 AC.a​

Answer 1

Answer:

Now is ∆ADB and ∆CDE we have  AD = DC [Given]  BD = DE [By construction]  And, ∠ADB = ∠CDE [Vertically opposite angles]  ∴ By SAS criterion of congruence we have  ∆ADB ≅ ∆CDE  ⇒ EC = AB and ∠CED = ∠ABD ....(i) [By cpctc]  But ∠CED & ∠ABD are alternate interior angles  ∴ CE ║ AB ⇒ ∠ABC + ∠ECB = 1800 [Consecutive interior angles] ⇒ 90 + ∠ECB = 1800  ⇒ ∠ECB = 900  Now, In ∆ABC & ∆ECB we have  AB = EC [By (i)]  BC = BC [Common]  And, ∠ABC = ∠ECB = 900  ∴ BY SAS criterion of congruence  ∆ABC ≅ ∆ECB  ⇒ AC = EB [By cpctc]  ⇒ 1/2 AC = 1/2 EB  ⇒ BD = 1/2 AC Read more on Sarthaks.com - https://www.sarthaks.com/75110/if-d-is-the-mid-point-of-the-hypotenuse-ac-of-a-right-triangle-abc-prove-that-bd-1-2-ac