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प्रश्न
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
- ΔAMC ≅ ΔBMD
- ∠DBC is a right angle.
- ΔDBC ≅ ΔACB
- CM = `1/2` AB
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उत्तर
Since M is the mid-point of AB.
∴ BM = AM
i. In ΔAMC and ΔBMD, we have
CM = DM ...[Given]
∠AMC = ∠BMD ...[Vertically opposite angles]
AM = BM ...[Proved above]
∴ ΔAMC ≅ ΔBMD ...[By SAS congruency]
ii. Since ΔAMC ≅ ΔBMD
∠MAC = ∠MBD ...[By corresponding parts of congruent triangles]
But they form a pair of alternate interior angles.
∴ AC ‖ DB
Now, BC is a transversal which intersects parallel lines AC and DB,
∴ ∠BCA + ∠DBC = 180° ...[Co-interior angles]
But ∠BCA = 90° ...[ΔABC is right angled at C]
∴ 90° + ∠DBC = 180°
⇒ ∠DBC = 90°
iii. Again, ΔAMC ≅ ΔBMD ...[Proved above]
∴ AC = BD ...[By corresponding parts of congruent triangles]
Now, in ΔDBC and ΔACB, we have
BD = CA ...[Proved above]
∠DBC = ∠ACB ...[Each 90°]
BC = CB ...[Common]
∴ ΔDBC ≅ ΔACB ...[By SAS congruency]
iv. As ΔDBC ≅ ΔACB
⇒ DC = AB ...[By corresponding parts of congruent triangles]
But DM = CM ...[Given]
∴ CM = `1/2` DC = `1/2` AB
⇒ CM = `1/2` AB.
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| Steps | Reasons | ||
| 1 | PM = QM | 1 | ... |
| 2 | ∠PMA = ∠QMA | 2 | ... |
| 3 | AM = AM | 3 | ... |
| 4 | ΔAMP ≅ ΔAMQ | 4 | ... |
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