Question

please answer this question ​

Answer 1

Answer:

a. In ΔPAC and ΔPDB

∠APC = ∠DPB        {Verically opposite angles}

∠PAC = ∠PBD        {Alternate angles}

∴ ΔPAC ≈ ΔPDB     {By A.A}

b. As ΔPAC ≈ ΔPDB

∴ PA/PD = PC/PB    {Correspondind sides of s similar Δ}

⇒ PA . PB = PC . PD

Hence Proved

Step-by-step explanation:

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Answer 2

Answer:

Give thanks to all my answers

Step-by-step explanation:

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