Question

12. In Fig. 7.126, AB = AC. Prove that AQ > AP.
A
P.
Q
B.
C
D​

Answer 1

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REF. Image.⬆

Given ,

• AB=AC
• and AP=AQ

Thus,

AB-AP=AC-AQ

[BP=CA ] [from figure ]

now InΔBCP & ΔBCQ

BP = CQ

∠c=∠c [common]

and BC=BC [common]

∴ΔBCP≃ΔBCQ [SAS congruency]

now,

• [BQ=CP] [corresponding parts of congruent triangles]

Answer 2

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REF. Image.⬆

Given ,

• AB=AC
• and AP=AQ

Thus,

AB-AP=AC-AQ

[BP=CA ] [from figure ]

now InΔBCP & ΔBCQ

BP = CQ

∠c=∠c [common]

and BC=BC [common]

∴ΔBCP≃ΔBCQ [SAS congruency]

now,

• [BQ=CP] [corresponding parts of congruent triangles]