Question

please solve this issue plzzzzzzzzz​

Answer 1

Given:-

• ABCD is a square AM = BM

To Prove:-

• PA = BQ

Solution:-

In Δ PAM and Δ QBM

⇢ ∠QBM = v.opposite to ∠PMA

⇢ AM = BM [given]

⇢ ∠MBQ + ∠MBC = 180° [linear pair]

⇢ ∠MBQ + 90° = 180° (as ABCD is a square)

⇢ ∠MBQ = 90°

⇢ ∠MAP = 90° (as ABCD is a square)

⇢ ∠MBQ = ∠MAP

∴ ΔPAM ≅ ΔQBM by A.S.A

∴ PA = BQ by CPCT.

Hence, proved!

Answer 2

Answer:

$\large \underline \mathfrak \red{Question}$

Given in the figure and asked to prove accordingly

$\large \underline \mathfrak \red{Answer}$

$\large \underline \green{ \rm \: Given}$

: ABCD is a SQUARE

→ so AB = BC = CD = AD

→ AM = BM

$\large \underline \green{ \rm \: To \: prove}$

: AP = BQ

$\large \underline \green{ \rm \:Proof }$

now compare ∆ APM and ∆ BMQ

<MBC = 90° ( angle of a square )

<MBQ = 180-90=90° ( straight line )

_________________________________

so now

in ∆ APM and ∆ BMQ

<A = < MBQ. ( proved above as 90° ) - i

AM = BM ( given ) - ii

<PMA = <BMQ ( vertically opposite angles ) - iii

___________________________________

so hence ∆ APM ≈ ∆ MBQ

so AP = BQ by C.P.C.T.C

$\large \underline \green{ \rm \: HENCE \: \: \: \: PROVED}$

$\rm \blue{@BrainlyBlockBusterBB}$