Question

please help very urgent​

Answer 1

Step-by-step explanation:

1) In ΔQLM and ΔRNM

QM = MR

LM = MN

∠QLM = ∠RNM = 90°

Therefore, ΔQLM ≅ ΔRNM ...(RHS criteria)

Hence, QL = RN ..........(i)

Join PM

In ΔPLM and ΔPNM and

PM = PM ...(common)

LM = MN

∠PLM = ∠PNM = 90°

Therefore, ΔPLM ≅ ΔPNM ...(RHS criteria)

Hence, PL = PN ..........(ii)

From (i) and (ii)

PQ = PR.

2) DP = CQ

⇒ DP + PQ = CQ + PQ

⇒ DQ = CP ...(1)

In right triangles ADQ and BCP,

∠ADQ = ∠BCP [each = 90°]

Hyp. AQ = Hyp. BP

Side DQ = Side CP

∴ ∆ADQ ≅ ∆BCP | RHS Axiom

∴ ∠DAQ = ∠CBP. | C.P.C.T.

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