Question

In Fig. 12.27, ABC is an isosceles triangle with AB = AC. P and Q
are points on AB and AC respectively such that AP = AQ.
(i) Is ∆ABQ=∆ACP
(ii) Is ∆BPC = ∆CQB ?
Give reasons in support of your answer.

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Answer 1

Answer:

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]

Step-by-step explanation:

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