Question

If D is the mid point of the hypotenuse AC of a right angle triangle ABC, prove that BD is equal to 1/2 AC?

Answer 1

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Answer 2

$\underline{ \underline{ { \huge{ \textbf{Q}}} \sf {UESTION - }}}$

If D is the midpoint of the hypotenuse AC of a right angle triangle ABC, prove that BD is equal to 1/2 AC.

$\underline{ \underline{ { \huge{ \textbf{G}}} \sf {IVEN -}}}$

• ABC is a right angled triangle.
• ∠ABC = 90°.
• D is the midpoint of AC.

$\underline{ \underline{{ \huge{ \textbf{T}}} \sf{O} \: \: \: { \huge{ \textbf{P}}} \sf{ROVE -}}}$

• BD = 1/2 AC.

$\underline{ \underline{ { \huge{ \textbf{C}}} \sf{ONSTRUCTION -}}}$

Draw DE parallel to CB which meets AB at point E.

$\underline{ \underline{ { \huge{ \textbf{P}}} \sf{ROOF -}}}$

since,

DE || CB and AB is transversal,

∠AED = ∠ABC = 90° = ∠DEB.

now,

since,

D is the midpoint of AC and DE || CB,

so,

DE bisects side AB [Converse mid point theorem]

so,

AE = BE.

Now,

In ∆ AED and ∆ BED,

∠AED = ∠BED [each 90°]

AE = BE [proved above]

DE = DE [common]

∴ ∆ AED ≅ ∆ BED [by S.A.S]

=> BD = AD [C.P.C.T]

=> BD = 1/2 AC [AD = 1/2 AC ]

_______________________proved)__

@MissSolitary✌️

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