In \(△ABC\), right-angled at \(C\), let \(X\) be the midpoint of the hypotenuse \(AB\). A line is drawn through \(X\), parallel to side \(BC\), meeting \(AC\) at \(D\). Prove the required result
Show that
(i) \(D\) is the mid-point of \(AC\)
(ii) \(MD ⊥ AC\)
(iii) \(CM = MA = \frac{1}{2}AB\)
Proof:
(i) To Prove: \(D\) is the mid-point of \(AC\)
Now, from \(∆ABC\) we can see \(X\) is the mid point of side \(AB\) and
\(DX || BC\) [Given]
The by Mid-point theorem, 'The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.'
Therefore, \(DC=\)
Hence, \(D\) is the mid-point of AC.
(ii) To Prove: \(XD ⊥ AC\)
Proof:
This implies, \(∠ACB = ∠ADX = ^°\)
Hence, \(XD ⊥AC\)
(i) To Prove: \(CX = AX = \frac{1}{2}AB\).
Proof:
Considering \(∆ADX\) and \(∆CDX\)
\(AD = \) [\(D\) is the mid point of \(AC\) ]
\(∠CDX = ∠\) (\(MD ⊥AC\))
\(DX = \) ()
\(∆ADX ≅ ∆CDX\) (By )
\(CX = \) (By )
\(CX = AX= \frac{1}{2} AB\) (X is the mid point of AB)