ABC triangle right angled at C. A line through the midpoint M of hypotenuse AB andparallel to BC intersects AC at D. (4)Show that:i) MD II BCii) CM = MA = ½ABiii) D is the midpoint of AC.
Question
ABC triangle right angled at C. A line through the midpoint M of hypotenuse AB andparallel to BC intersects AC at D. (4)Show that:i) MD II BCii) CM = MA = ½ABiii) D is the midpoint of AC.
Solution
i) MD || BC: Since the line through M is given as parallel to BC, MD is also parallel to BC by definition of parallel lines.
ii) CM = MA = ½AB: Since M is the midpoint of AB, by definition, CM = MA. And since M is the midpoint, each of these segments is half the length of AB, so CM = MA = ½AB.
iii) D is the midpoint of AC: Since MD is parallel to BC and AD is a transversal, then by the Alternate Interior Angles Theorem, ∠MAD = ∠ABC. Similarly, ∠ADM = ∠ACB. Therefore, triangle ADM is similar to triangle ABC (by the AA postulate). Since AB = 2AM (from part ii), then AC = 2AD by the properties of similar triangles, which means D is the midpoint of AC.
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