Question

14. In the given figure, ABCD is a square and APAB is an equilateral
triangle
(i) Prove that AAPD = ABPC.
(ii) Show that DPC = 15°.
Hint. (i) _PAD = 90° + 60° = 150° and ZPBC = 90° + 60° = 150°.
(ii) _PAD = 150° and AP = AD = LAPD = ZADP = 15°.]​

Answer 1

Answer:

In △ABP,

AP=BP [Sides of equilateral △]

& AD=BC [Sides of equilateral △]

and ∠DAP=∠DAB−∠PAB=90

∘

−60

∘

=30

∘

Now in △APD and △BPC

AP=BP [Sides of equilateral △]

AD=BC [Sides of equilateral △]

∠DAP=∠CBP [Both 30

∘

]

∴△APD≅△BPC [By SAS]

In △APD

AP=AD [∵ AP=AB ]

∠DAP=30

∘

∠APD=

2

180

∘

−30

∘

=75

∘

Similarly, ∠BPC=75

∘

Therefore, ∠DPC=360

∘

−(75+75+60)=150

∘