Question

hey koi help kr do plz ​

Answer 1

Solution:

$\sf \colorbox{pink} {(i) In ∆ABD and ∆ACD}$

$\sf \colorbox{god} {AB = AC (∴∆ABC is isosceles}$

$\sf \colorbox{god} {AD = AD (common)}$

$\sf \colorbox{god} {RD = DC (∆ABC is isosceles)}$

$\sf \colorbox{god} {∆ABD ≈ ∆ACD (sss- congruently)}$

$\sf \colorbox{god} {∴∠ABD ≈ ∠CAD i.e. ∠BAP = ∠PAC ...(i) (CPCT) }$

$\sf \colorbox{pink} {(ii) In ∆ABP = ∆ACP }$

$\sf \colorbox{god} {∴AB = AC (∆ABC is isosceles)}$

$\sf \colorbox{god} {AP = AP (common side)}$

$\sf \colorbox{god} {∠BAP = ∠PAC (from (i) )}$

$\sf \colorbox{god} { ∆ABP ≈ ∆ACP (SAS ingraining)}$

$\sf \colorbox{god} {∴BP = PC (ii) (CPCT )}$

$\sf \colorbox{god} {∠APB = ∠APC (CPCT)}$

$\sf \colorbox{pink} {(iii) AP bisects ∠A as well as ∠D}$

$\sf \colorbox{god} {∴(ii) ∆ABD ≈ ∆ACD}$

$\sf \colorbox{god} {∠BAD = ∠CAD (from (i) )}$

$\sf \colorbox{god} {so AD c = bisect, ∠A, i.e. AP bisect ∠A ...(iii)}$

$\sf \colorbox{god} {In ∆BDP and ∆CDP}$

$\sf \colorbox{god} {OP = BP (common)}$

$\sf \colorbox{god} {BP = PC (from (ii) )}$

$\sf \colorbox{god} {BD = CD ∆BDC isosceles}$

$\sf \colorbox{god} {∆BDP ≈ ∠CDP (SS contravening)}$

$\sf \colorbox{god} {∴∠BDP = ∠CDP (CPCT)}$

‣DP Bisect ∠D, so AP bisect ∠A as well as ∠D

$\sf \colorbox{pink} {(iv) ∠APB + ∠APC = 180° ( linear pair)}$

$⇒∠APB = ∠APC = \frac{180}{2} = 90$

$\sf \colorbox{god} {BP =PC and ∠APB = ∠APC = 90}$

Hence,

$\sf \colorbox{god} {AP is Perpendicular Bisector of BC}$

$\\ \\ \\ \\ \sf \colorbox{lightgreen} {\red★ANSWER ᵇʸɴᴀᴡᴀʙ}$