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प्रश्न
In the given figure, prove that:
CD + DA + AB > BC

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उत्तर
We have to prove that CD + DA + AB > BC
In ΔACD we have
CD + DA > CA (As sum of two sides of triangle is greater than third one)
⇒ CD + DA + AB > CA + AB (Adding AB both sides)
CD + DA + AB > BC Proved.
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संबंधित प्रश्न
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:
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You have to show that ΔAMP ≅ AMQ.
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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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