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प्रश्न
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that
- ΔDAP ≅ ΔEBP
- AD = BE
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उत्तर
We have, P is the mid-point of AB.
∴ AP =BP
∠EPA = ∠DPB ...[Given]
Adding ∠EPD to both sides, we get:
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒ ∠APD = ∠BPE
i. Now, in △DAP and △EBP, we have
∠PAD = ∠PBE ...[∵ ∠BAD = ∠ABE]
AP = BP ...[Proved above]
∠DPA = ∠EPB ...[Proved above]
∴ △DAP ≌ △EBP ...[By ASA congruency]
ii. Since △DAP ≌ △EBP
⇒ AD = BE ...[By Corresponding Parts of Congruent Triangles]
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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:
- ΔAPB ≅ ΔAQB
- BP = BQ or B is equidistant from the arms of ∠A.
You have to show that ΔAMP ≅ AMQ.
In the following proof, supply the missing reasons.
| Steps | Reasons | ||
| 1 | PM = QM | 1 | ... |
| 2 | ∠PMA = ∠QMA | 2 | ... |
| 3 | AM = AM | 3 | ... |
| 4 | ΔAMP ≅ ΔAMQ | 4 | ... |
If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.
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(ii) AB = CE.
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