Question

Plzz solve question 14!!!

Answer 1

Ad=Bc
<bac=<PAC
PA=PD
So tri<APD=Tri<BPC
So, PA=PC
Which is possible when line in square intersect at 90°

Answer 2

$\huge{\pink{\underline{\ulcorner{\star\: Solution\: \star}\urcorner}}}$

$\red{\underline{Given \: \colon}}$

(a) ABCD is a square and P is any point in the square.

(b) PB = PD

$\red{\underline{To\: Prove\:\colon}}$

CPA is a straight line

$\red{\underline{Solution}}$

ABCD is a square . So,

AB = BC = CD = DA

$\red{\huge{\underline{Part\: A\: \colon}}}$

$In\: \triangle{APD} \& \: \triangle{APB} \\\\ AD = AB ...........( side of square)\\\\ DP = PB ...........( given) \\\\ AP = AP .............(common)$

Now,

$\triangle{APD} \: \cong \: \triangle{APB}$

By SSS - Congruency rule

$\blue{\boxed{\angle{1} = \angle{2}}}$..(1)

By C.P.C.T { Corresponding part of congruent Triangle }

$\huge{\red{\underline{Part\: B\: \colon }}}$

$In\: \triangle{DPC}\: \& \: \triangle{CPB}\\\\ DP = PB ............(given) \\\\ DC = CB ..............(side of square) \\\\ CP = CP ................(common)$

Now,

$\triangle{DPC}\: \cong \: \triangle{CPB}$

By SSS - Congruency rule

$\blue{\boxed{\angle{3} = \angle{4}}}$..(2)

By C.P.C.T { Corresponding part of congruent triangle }

Now

add eq 1 & 2

$\angle{1}+\angle{3} = \angle{2}+\angle{4}$

.....(3)

Now sum of all angles

$\angle{1}+\angle{2}+ \angle{3}+ \angle{4} = 360\degree \\\\ (\angle{1}+\angle{3})+(\angle{2} + \angle{4}) = 360\degree \\\\ \therefore \: put \: the \: value\: in \: equation \: from \: (3) \\\\ (\angle{1}+\angle{3}) +(\angle{1}+\angle{3}) = 360\degree\\\\ 2(\angle{1}+\angle{3})=360\degree\\\\ \angle{1} + \angle{3} = 180\degree$

Means these angles are in linear pair and a line CAP is striaght line.

Thanks