Maths problem: how to do, thanks?

Question：
In the figure, ABC is a triangle and D, P and E lie on BC. It is given that AP = 8 cm, BD = EC = 9 cm and ∠ABC = ∠ACB.
(a) Show that ΔABE ≅ ΔACD.
(b) It is given that DP = EP = 6 cm.
(i) Show that ΔACP is a right-angled triangle.
(ii) Hence find AC.
(c) Is  ΔACD a right-angled triangle? Explain your answer.

https://s.yimg.com/tr/i/c176320ffce34dc0b800358aa9b18898_A.png
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Solution：
(a)
In ΔABE and ΔACD,
BE = BD + DE
      = CE + DE ...... ( given )
      = CD
∠ABE =  ∠ACD ...... ( given )
AB = AC ...... ( sides opp. eq. ∠s )
∴ ΔABE ≅ ΔACD ...... ( SAS )

(bi)
In ΔABP and ΔACP,
AB = AC ...... ( proved )
BP = BD + DP
      = CE + EP ...... ( given )
      = CP
AP = AP ...... ( common side )
∴ ΔABE ≅ ΔACD ...... ( SSS )
∴ ∠APB = ∠APC  ...... ( corr. ∠s, ≅ Δs )
∠APC = (∠APB + ∠APC)/2 = 180°/2 = 90° ...... ( adj. ∠s on st. li. )
∴ ΔACP is a right-angled triangle.
(bii)
AC
= √(AP² + CP²) ...... ( Pyth. theorem )
= √[8² + (9 + 6)²]
= 17 

(c)
AC² + AD²
= AC² + AP² + DP² ...... ( Pyth. theorem )
= 17² + 8² + 6²
= 389
≠ (9 + 6 + 6)² = CD²
∴ ΔACD is not a right-angled triangle.
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https://yahoo.uservoice.com/forums/924010-yahoo-answers/suggestions/40973842-bring-back-comments
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謝謝你,原來要利用畢氏定理才計算得到.我想了很久也想不通.現在終於明白了