Question

hii everyone please solve my questions.​

Answer 1

$\huge{\boxed{\mathfrak{\underline{Answer:}}}}$

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$\large{\sf{Mathematics}}$

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$\bf Given:$

ΔABC and ΔDBC are two isosceles triangles in which AB=AC & BD=DC.

$\bf To \: Prove:$

(i) ΔABD ≅ ΔACD

(ii) ΔABP ≅ ΔACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC.

$\bf Proof:$

(i) In ΔABD and ΔACD,

AD = AD (Common)

AB = AC (given) .

BD = CD (given)

Therefore, ΔABD ≅ ΔACD (by SSS congruence rule)

∠BAD = ∠CAD (CPCT)

∠BAP = ∠CAP

(ii) In ΔABP & ΔACP,

AP = AP (Common)

∠BAP = ∠CAP

(Proved above)

AB = AC (given)

Therefore,

ΔABP ≅ ΔACP

(by SAS congruence rule).

(iii) ∠BAD = ∠CAD (proved in part i)

Hence, AP bisects ∠A.

also,

In ΔBPD and ΔCPD

PD = PD (Common)

BD = CD (given)

BP = CP (ΔABP ≅ ΔACP so by CPCT)

Therefore, ΔBPD ≅ ΔCPD (by SSS congruence rule)

Thus,

∠BDP = ∠CDP( by CPCT)

Hence, we can say that AP bisects ∠A as well as ∠D.

(iv) ∠BPD = ∠CPD

(by CPCT as ΔBPD ≅ ΔCPD)

& BP = CP (CPCT)

∠BPD + ∠CPD = 180° (BC is a straight line)

⇒ 2∠BPD = 180°

⇒ ∠BPD = 90°

Hence, AP is the perpendicular bisector of BC.

Answer 2

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here is your answer....

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