Question

A 1. AABC and A DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that (1) A ABD=AACD () A ABPSA ACP Let us ual an B P (ii) AP bisects LA as well as D. (iv) AP is the perpendicular bisector of BC. vity BC Fig. 7.39 Fix​

Answer 1

$\: \huge\sf\blue{➳\: Correct \: Question : - }$

➳△ABC and △DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that

(i) △ABD≅△ACD

(ii) △ABP≅△ACP

(iii) AP bisects ∠A as well as △D.

(iv) AP is the perpendicular bisector of BC.

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$\: \: \: \: \: \: \: \huge\sf\blue{⇝Sᴏʟᴜᴛɪᴏɴ : - }$

(i) In △ABD and △ACD,

AB=AC ⇝ (since △ABC is isosceles)

AD=AD ⇝ (common side)

BD=DC ⇝ (since △BDC is isosceles)

ΔABD≅ΔACD ⇝SSS test of congruence,

∴∠BAD=∠CAD i.e. ∠BAP=∠PAC

[c.a.c.t]⇝(i)

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(ii) In △ABP and △ACP,

AB=AC ⇝ (since △ABC is isosceles)

AP=AP ⇝ (common side)

∠BAP=∠PAC ⇝from (i)

△ABP≅△ACP ⇝SAS test of congruence

∴BP=PC [c.s.c.t]⇝(ii)

∠APB=∠APC ⇝c.a.c.t.

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(iii) Since △ABD≅△ACD

∠BAD=∠CAD ⇝from (i)

So, AD bisects ∠A

i.e. AP bisects∠A⇝(iii)

In △BDP and △CDP,

DP=DP ⇝ (common side)

BP=PC ⇝from (ii)

BD=CD (since △BDC is isosceles)

△BDP≅△CDP ⇝SSS test of congruence

∴∠BDP=∠CDP ⇝c.a.c.t.

∴ DP bisects∠D

So, AP bisects ∠D ⇝(iv)

From (iii) and (iv),

AP bisects ∠A as well as ∠D.

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(iv) We know that

∠APB+∠APC=180°⇝(angles in linear pair)

Also, ∠APB=∠APC ⇝from (ii)

∴∠APB=∠APC= 180°/2 = 90°

⠀⠀⠀⠀⠀⠀

BP=PC and ∠APB=∠APC=90°

$\bf{Hence, \: AP \: is \: perpendicular \: bisector \: of \: BC.}$

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