Smile 發問時間： 科學及數學數學 · 1 個月前

# Maths problem: how to do, thanks?

### 2 個解答

• 匿名使用者
1 個月前
最佳解答

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.

https://s.yimg.com/tr/i/c176320ffce34dc0b800358aa9...

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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 + 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-an...

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• 1 個月前

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